Materi kelas X, Materi kelas X semester I, Pertidaksamaan, Uncategorized

Pertidaksamaan

A. Pengertian

Secara definisi

Pertidaksamaan adalah kalimat terbuka dalam matematika yang dihubungkan dengan tanda \left ( < ,\: \leq ,\: > ,\:\: \textrm{dan}\: \: \geq \right ).

Penjelasan lebih lanjut

\begin{array}{|c|c|l|}\hline \textrm{Notasi}&\textrm{Ilustrasi}&\textrm{Dibaca}\\\hline < &m< n&\textrm{\textit{m} kurang dari \textit{n}}\\\hline \leq &m\leq n&\textrm{\textit{m} kurang dari atau sama dengan \textit{n}}\\\hline > &m> n&\textrm{\textit{m} lebih dari \textit{n}}\\\hline \geq &m\geq n&\textrm{\textit{m} lebih dari atau sama dengan \textit{n}}\\\hline \multicolumn{3}{|c|}{\textrm{Sebagai tambahan penjelasan}}\\\hline \multicolumn{2}{|c|}{m< x< n}&\textrm{\textit{x} lebih dari \textit{m} dan \textit{x} kurang dari \textit{n}}\\\hline \multicolumn{2}{|c|}{m\leq x\leq n}&\textrm{menyesuaikan}\\\hline \multicolumn{2}{|c|}{m< x\: \: \textrm{atau}\: \: x> n}&\textrm{\textit{x} kurang dari \textit{m} atau \textit{x} lebih dari \textit{n}}\\\hline \multicolumn{2}{|c|}{m\leq x\: \: \textrm{atau}\: \: x\geq n}&\textrm{Menyesuaikan}\\\hline \end{array}.

Sifat Operasi pertidaksamaan

\begin{array}{|l|l|l|l|l|l|}\hline &\multicolumn{5}{c|}{\textrm{Operasi Pertidaksamaan untuk a, b, c bilangan real}}\\\cline{2-6}\raisebox{1.5ex}[0cm][0cm]{No}&\textrm{Urutan}&\textrm{Dijumlah/dikurangi}&\textrm{Dikalikan/dibagi}&\textrm{Dikalikan/dibagi}&\textrm{Dikuadratkan}\\\hline 1.&\begin{aligned}&a< b\\ &b< c\\ &\textrm{maka}\\ &a< c\\ &\\ &\\ & \end{aligned}&\begin{aligned}&a< b\\ &\textrm{maka berlaku}\\ &a+c< c+b\\ &\textrm{atau}\\ &a-c< b-c\\ &\\ & \end{aligned}&\begin{aligned}&\textrm{tidak berubah}\\ &\textrm{tanda jika c}\\ &\textrm{bilangan positif}\\ &a< b\\ &\textrm{maka}\\ &ac< bc\\ & \end{aligned}&\begin{aligned}&\textrm{berubah tanda}\\ &\textrm{jika c negatif}\\ &a< b\\ &\textrm{maka}\\ &ac> bc\\ &\textrm{demikian juga}\\ &\textrm{sebaliknya} \end{aligned}&\begin{aligned}&a< b\\ &\textrm{maka}\\ &a^{2}< b^{2}\\ &\textrm{atau}\\ &a> b\\ &\textrm{maka}\\ &a^{2}> b^{2} \end{aligned}\\\hline 2.&\multicolumn{5}{l|}{\begin{aligned}&\textrm{Diketahui}\\ &\displaystyle \frac{a}{c}< \displaystyle \frac{b}{c}\\ &\textrm{maka jika dibalik bilangan pecahan ini akan menjadi}\\ &\displaystyle \frac{c}{a}> \displaystyle \frac{c}{b} \end{aligned}}\\\hline \end{array}.

B. Pertidaksamaan Linear

Pertidaksamaan linear didefinisikan sebagai pertidaksamaan linear untuk variabel dengan pangkat tertinggi satu.

\LARGE{\fbox{\LARGE{\fbox{Contoh Soal}}}}.

\begin{array}{ll}\\ \fbox{1}.&\textrm{Tentukanlah himpunan penyelesaian dari}\\ &\begin{array}{lllllllll}\\ a.\quad x+3> 0&f.\quad 2x-5\leq 6x+3\\ b.\quad 3x+18\leq 0&g.\quad 1-x< 2+7x\\ c.\quad 14+7x\geq 0&h.\quad \displaystyle \frac{1}{6}x-5\geq \displaystyle \frac{1}{2}x+\frac{1}{4}\\ d.\quad 6\leq -2x&i.\quad \displaystyle \frac{1}{2}x-\frac{5}{12}< -\frac{1}{3}x+\frac{3}{2}\\ e.\quad -2x-8< 0&j.\quad -\displaystyle \frac{1}{2}x+\frac{7}{8}x> \displaystyle \frac{1}{4}x-\frac{5}{8}\end{array} \end{array}.

\textrm{Jawab}:\\\\ \begin{array}{llllll}\\ \begin{aligned}\textrm{a}.\quad x+3&> 0\\ x&> -3\\ &\\ \textrm{HP}=&\left \{ x|x> -3,\: x\in \mathbb{R} \right \} \end{aligned}&\quad \begin{aligned}\textrm{b}.\quad 3x+18&\leq 0\\ x+6&\leq 0\quad \textrm{(masing-masing ruas dibagi 3)}\\ x&\leq -6\\ \textrm{HP}=&\left \{ x |x\leq -6,\: x\in \mathbb{R}\right \} \end{aligned}&&&&\\\\ \begin{aligned}\textrm{c}.\quad 14+7x&\geq 0\\ 2+x&\geq 0\\ x&\geq -2\\ &\\ \textrm{HP}=&\left \{ x|x\geq -2,\: x\in \mathbb{R} \right \}\\ &\end{aligned}&\quad \begin{aligned}\textrm{d}.\quad 6&\leq -2x\\ -2x&\geq 6\\ -x&\geq 3\\ x&\leq -3\quad \textrm{(perhatikanlah perubahan tandanya)}\\ &\\ \textrm{HP}=&\left \{ x|x\leq -3,\: x\in \mathbb{R} \right \} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{2}.&\textbf{(UMPTN 1996)}\\ &\textrm{Pertidaksamaan}\: 2x-a\: > \displaystyle \frac{x-1}{2}+\displaystyle \frac{ax}{3}\: \textrm{memiliki penyelesaian}\: \: x> 5\\ &\textrm{Nilai dari a adalah ....} \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{aligned}2x-a\: &\displaystyle > \frac{x-1}{2}+\displaystyle \frac{ax}{3}\quad \textrm{(masing-masing ruas dikali 6)}\\ 12x-6a\: &\displaystyle > 3x-3+2ax\\ 12x-3x-2ax\: &> 6a-3\\ \left ( 9-2a \right )\: &> 6a-3\\ x\: &> \displaystyle \frac{6a-3}{9-2a},\quad \textrm{padahal penyelesaiannya adalah} \: x> 5\\ \textrm{maka}&\\ \displaystyle \frac{6a-3}{9-2a}&=5\\ 6a-3&=45-10a\\ 6a+10a&=45+3\\ 16a&=48\\ a&=3\\ \end{aligned}\\\\ \textrm{Jadi, nilai a adalah 3}.

\begin{array}{ll}\\ \fbox{3}.&\textbf{(UMPTN 1994)}\\ &\textrm{Jika} \: \: a< x< b,\: \textrm{dan}\: \: a< y< b,\: \: \textrm{maka nilai }x-y \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{aligned}\textrm{Diketahui bahwa}&\\ a< &x< b\: \: .......\textcircled{1}\\ a< &y< b\: \: .......\textcircled{2}&\textrm{untuk yang ini dikalikan (-1), sehingga menjadi}\\ &&a<x<b \: \: \: \: \: \: \: \: \: \\ &&-b<-y<-a \: \: \: \: \\ &&------\: \: +\\ &&a-b<x-y< b-a\\ \end{aligned}\\\\ \textrm{Jadi, harga} \: x-y\: \textrm{adalah ada pada rentang}\: a-b<x-y< b-a ..

\begin{array}{lp{18.0cm}}\\ \fbox{4}.&\textbf{(UMPTN 1997)}\\ &\textrm{Diketahui P, Q, dan R memancing ikan. Jika hasil Q lebih sedikit dari hasil R sedangkan jumlah hasil P dan Q lebih banyak dari pada dua kali hasil R, maka yang terbanyak mendapat ikan adalah ....}\\ &a. P dan R\qquad\quad\quad \: \: \: \: \: \: \: \: \, d. Q\\ &b. P dan Q\qquad \: \: \: c. P\qquad \, \, \, e. R \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{aligned}\textrm{Diketahui:}&\: \: \: \\ &\bullet \: Q< R\: ...............\textcircled{1}\\ &\bullet \: P+Q> 2R\: ......\textcircled{2}\\ &\textrm{Sehingga}\\ &&R> Q\: \: \: \: \: \: \: \\ &&P+Q> 2R\: \: \: \: \: \\ &&-----\: \: \: \: +\\ &&P+Q+R> Q+R+R\\ &&P> R\: \: .........\textcircled{3}\\ &\textrm{dari} \: \textcircled{1}\: \textrm{dan}\: \textcircled{3}\: \textrm{diperoleh bahwa}\\ &Q<R< P \end{aligned}\\\\ \textrm{Jadi, yang terbanyak mendapat ikan adalah P}.

\begin{array}{ll}\\ \fbox{5}.&\textrm{Tentukanlah bilangan bulat terbesar n sehingga ada bilangan unik k yang memenuhi} \\ &\displaystyle \frac{8}{15}< \displaystyle \frac{n}{n+k}< \displaystyle \frac{7}{13}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\left (\textrm{American Invitational Mathematics Examinations 1987} \right )\\ \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{aligned}\textrm{Diketahui}&\: \textrm{bahwa}\\ &\displaystyle \frac{8}{15}< \displaystyle \frac{n}{n+k}< \displaystyle \frac{7}{13}\quad (\textrm{jika dibalik, maka akan menjadi})\\ &\displaystyle \frac{15}{8}> \displaystyle \frac{n+k}{n}> \displaystyle \frac{13}{7}\\ &\textrm{atau}\\ &\displaystyle \frac{13}{7}< \displaystyle \frac{n+k}{n}< \displaystyle \frac{15}{8}\quad (\textrm{samakan penyebutnya yaitu dengan})\\ &\displaystyle \frac{13}{7}\: .\: \frac{8}{8}< \displaystyle \frac{n+k}{n}< \displaystyle \frac{15}{8}\: .\: \frac{7}{7}\\ &\displaystyle \frac{104}{56}< \displaystyle \frac{n+k}{n}< \displaystyle \frac{105}{56}\quad (\textrm{masing-masing dikalikan dua})\\ &\displaystyle \frac{208}{112}< \displaystyle \frac{n+k}{n}< \displaystyle \frac{210}{112}\quad (\textrm{pertidaksamaan akan urut saat n+k = 209 dengan n = 112}) \end{aligned}\\\\ \textrm{Jadi, nilai n = 112}.

\begin{array}{ll}\\ \fbox{7}.&\textrm{Tentukanlah himpunan penyelesaian dari}\\ &\begin{array}{lllllllll}\\ \textrm{a}.\quad -8< 2x-3\leq 5&\textrm{c}.\quad 3x+4\leq 5x+6< 2x+12\\ \textrm{b}.\quad 9-x< 3x-7\leq 4x-4&\textrm{d}.\quad 2x+2\leq x-2\leq 2x-3\\ \end{array} \end{array}\\\\\\ \textrm{Jawab}:\\ \begin{array}{lll}\\ \begin{aligned}a.\quad &-8< 2x-3\leq 5\\ -8&+3<2x-3+3\leq 5+3\\ &-5< 2x\leq 8\\ &\displaystyle -\frac{5}{2}< x\leq 4\\ &\\ &\\ \textrm{HP}=&\left \{x|-\displaystyle \frac{5}{2}< x\leq 4,\: x\in \mathbb{R} \right \} \end{aligned}&\begin{aligned}b.\qquad &\underset{\overbrace{\begin{matrix} \begin{aligned}5x+6&\geq 3x+4\\ 5x-3x&\geq 4-6\\ 2x&\geq -2\\ x&\geq -1 \end{aligned} & \begin{aligned}5x+6&<2x+12\\ 5x-2x&< 12-6\\ 3x&< 6\\ x&< 2 \end{aligned} \end{matrix}}}{3x+4\leq 5x+6< 2x+12}\\ &\\ \textrm{HP}=&\left \{ x|-1\leq x< 2,\: x\in \mathbb{R} \right \}\\ &\\ & \end{aligned}& \end{array}.

C. Pertidakamaan Kuadrat

Bentuk Umum :

\begin{aligned}ax^{2}+bx+c&\: \: \begin{cases} < \\ \leq \\ > \\ \geq \end{cases} 0\\ \textrm{dengan}\: &a, \: b,\: c\: \in \mathbb{R},\: a\neq 0 \end{aligned}..

\begin{array}{|c|c|c|c|}\hline \textrm{Pertidaksamaan Kuadrat}&\textrm{Akar}&\textrm{Syarat}&\textrm{Himpunan Penyelesaian (HP)}\\\hline \begin{aligned}ax^{2}+bx+c&\: \: \begin{cases} < \\ \leq \\ > \\ \geq \end{cases} 0\\ \textrm{dengan}\: &a, \: b,\: c\: \in \mathbb{R},\: a\neq 0 \end{aligned} &\begin{aligned}&x_{1}\: \textrm{dan}\: x_{2}\\ &\textrm{dengan}\: x_{1}\neq x_{2} \end{aligned}&x_{1}< x_{2}&\begin{cases} ax^{2}+bx+c< 0, &\textrm{HP}=\left \{ x|x_{1}< x< x_{2},\: x\in \mathbb{R} \right \} \\ ax^{2}+bx+c\leq 0, &\textrm{HP}=\left \{ x|x_{1}\leq x\leq x_{2},\: x\in \mathbb{R} \right \} \\ ax^{2}+bx+c> 0, &\textrm{HP}=\left \{ x|x< x_{1}\: \textrm{atau}\: x> x_{2},\: x\in \mathbb{R} \right \} \\ ax^{2}+bx+c\geq 0, &\textrm{HP}=\left \{ x|x\leq x_{1}\: \textrm{atau}\: x\geq x_{2},\: x\in \mathbb{R} \right \} \end{cases}\\\hline \end{array}.

\LARGE{\fbox{\LARGE{\fbox{Contoh Soal}}}}.

\begin{array}{ll}\\ &\textrm{Tentukanlah himpunan penyelesaian dari}\\ &\begin{array}{lllllllll}\\ \textrm{a}.\quad x^{2}-2x-8< 0&\textrm{c}.\quad 3x^{2}-2x+5> 0\\ \textrm{b}.\quad -x-x^{2}< -6&\textrm{d}.\quad -x^{2}+x-3\leq 0\\ \end{array} \end{array}\\\\\\ \textrm{Jawab}:\\ \begin{array}{lllll}\\ \begin{aligned}\textrm{a}.\quad &x^{2}-2x-8< 0\\ &\textrm{ubahlah tanda pertidaksamaan}\\ &\textrm{dengan persamaan, sehingga}\\ &x^{2}-2x-8=0\\ &(x-4)(x+2)=0\\ &x=4\: \: \textrm{atau}\: \: x=-2\\ &\underset{\begin{matrix} |\\ x_{1} \end{matrix}}{x=-2}\: \: \textrm{atau}\: \: \underset{\begin{matrix} |\\ x_{2} \end{matrix}}{x=4}\\ &\textrm{HP}=\left \{ x|-2< x< 4,\: x\in \mathbb{R} \right \} \end{aligned}&\begin{aligned}\textrm{b}.\quad &-x-x^{2}< -6 \Rightarrow 6-x-x^{2}< 0\Rightarrow x^{2}+x-6> 0\quad (\textrm{dikalikan dengan -1})\\ &\textrm{ubahlah tanda pertidaksamaan}\\ &\textrm{dengan persamaan, sehingga}\\ &x^{2}+x-6=0\\ &(x+3)(x-2)=0\\ &x=-3\: \: \textrm{atau}\: \: x=2\\ &\underset{\begin{matrix} |\\ x_{1} \end{matrix}}{x=-3}\: \: \textrm{atau}\: \: \underset{\begin{matrix} |\\ x_{2} \end{matrix}}{x=2}\\ &\textrm{HP}=\left \{ x|x< -3\: \textrm{atau}\: x> 2,\: x\in \mathbb{R} \right \} \end{aligned}&& \end{array}.

Untuk jawaban c dan d karena keduanya masing-masing adalah definit positif (c) dan definit negatif (d), maka  \textrm{HP}=\left \{ x|x\in \mathbb{R} \right \}.

D. Pertidaksamaan Bentuk Pecahan

Pertidaksamaan bentuk pecahan adalah pertidaksamaan yang memuat pecahan bentuk aljabar.

\begin{array}{|c|c|c|c|}\hline \textrm{Bentuk Pertidaksamaan}&\textrm{Perlu diingat}&\textrm{Misal}&\textrm{Untuk Bagian Penyebut}\\\hline \displaystyle \frac{f(x)}{g(x)}\: \: \begin{cases} < \\ \leq \\ > \\ \geq \end{cases}\displaystyle \frac{h(x)}{k(x)}&\begin{aligned}&\textrm{Tidak boleh dikali silang}\\ &\textrm{jika penyebutnya baik}\\ &\textrm{salah satu atau keduanya}\\ &\textrm{mengandung variabel x} \end{aligned}&\displaystyle \frac{6}{2x-8}< \frac{3x}{2x-8}&\begin{aligned}&\textrm{Jika pada pertiaksamaan bertanda}\\ &\displaystyle \frac{\cdots }{x-a}\geq \frac{\cdots }{x-b}\: \: \textrm{atau}\: \: \displaystyle \frac{\cdots }{x-a}\leq \frac{\cdots }{x-b}\\ &\textrm{maka}\: x\neq a\: \textrm{dan}\: x\neq b\end{aligned}\\\hline \end{array} .

\LARGE{\fbox{\LARGE{\fbox{Contoh Soal}}}}.

\begin{array}{ll}\\ \fbox{1}.&\textrm{Tentukan himpunan penyelesaian dari pertidaksamaan}\\ &\begin{array}{ll}\\ \textrm{a}.\quad \displaystyle \frac{1}{x}\geq 2&\qquad \textrm{c}.\quad \displaystyle \frac{3x-2}{x}< x\\ \textrm{b}.\quad \displaystyle \frac{6}{2x-8}< \displaystyle \frac{3x}{2x-8}&\qquad \textrm{d}.\quad \displaystyle \frac{x-1}{x-2}\geq \frac{x-3}{x-4} \end{array} \end{array} .

Jawab:

\begin{array}{l}\\ \begin{aligned}\textrm{a}.\quad & \displaystyle \frac{1}{x}\geq 2\\ &\displaystyle \frac{1}{2}-2\geq 0\quad (\textrm{tidak boleh dikali silang})\\ &\displaystyle \frac{1}{x}-\frac{2x}{x}\geq 0\quad (\textrm{samakan penyebutnya})\\ &\displaystyle \frac{1-2x}{x}\geq 0\quad (\textrm{koefisien x pada pembilang dikali -1, untuk memudahkan saja})\\ &\displaystyle \frac{2x-1}{x}\leq 0\quad \textrm{boleh dipandang sebagai}\quad \displaystyle \frac{\left ( x-\frac{1}{2} \right )}{(x-0)}\leq 0\\ &\textrm{HP}=\left \{ x|0<x\leq \displaystyle \frac{1}{2},\: x\in \mathbb{R} \right \} \end{aligned}\end{array} .

Dengan garis bilangan

331

\begin{array}{l}\\ \begin{aligned}\textrm{b}.\quad & \displaystyle \frac{6}{2x-8x}< \frac{3x}{2x-8}\\ &\displaystyle \frac{6}{2x-8}-\frac{3x}{2x-8}< 0\\ &\displaystyle \frac{6-3x}{2x-8}< 0\quad (\textrm{koefisien x pada pembilang dikali -1, untuk memudahkan saja dan tanda berubah})\\ &\displaystyle \frac{3x-6}{2x-8}> 0\quad \textrm{boleh dipandang sebagai}\quad \displaystyle \frac{\left ( x-\frac{6}{3} \right )}{(x-\frac{8}{2})}> 0\: \: \textrm{atau}\: \: \displaystyle \frac{\left (x-2 \right )}{\left (x-4 \right )}> 0\\ &\textrm{HP}=\left \{ x|x< 2\: \textrm{atau}\: x> 4,\: x\in \mathbb{R} \right \} \end{aligned}\end{array} .

Dengan garis bilangan

332

\begin{array}{ll}\\ \fbox{2}.&\textrm{Tentukan himpunan penyelesaian dari pertidaksamaan}\\ &\begin{array}{ll}\\ \textrm{a}.\quad \displaystyle \frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-6)}\geq 0&\qquad \textrm{b}.\quad \displaystyle \frac{(x+1)(x-1)^{2}(x-5)^{3}}{(x+2)(x-2)(x-3)}\leq 0 \end{array} \end{array} .

Jawab:

Untuk jawaban a. perhatikanlah garis bilangan berikut

333

atau Anda dapat menggunakan titik uji ambil x = 10, masukkan ke pertidaksamaan dan hasilnya akan menunjukkan hal yang sama.

Jadi,  \textrm{HP}=\left \{ x\: |\: x\leq 1\: \textrm{atau}\: 2\leq x\leq 3\: \textrm{atau}\: 4< x< 5\: \textrm{atau}\: \: x> 6,\: x\in \mathbb{R} \right \} .

Untuk jawaban b. kita dihadapkan dengan bagian perpangkatan. Untuk faktor yang berpangkat genap maka sekitarnya adalah bertanda sama kalau ganjil tidak. Perhatikan ilustrasi berikut

335

Jadi,  \textrm{HP}=\left \{ x\: |\: x< -2\: \textrm{atau}\: -1\leq x< 2\: \textrm{atau}\: 3< x\leq 5\: ,\: x\in \mathbb{R} \right \} .

Catatan : penyebut tidak boleh nol. sehingga faktor yang posisinya berada di penyebut tandanya hanya berupa  <  dan  >  saja.

\begin{array}{ll}\\ \fbox{3}.&\textrm{Tentukan himpunan penyelesaian dari pertidaksamaan}\\ &\begin{array}{ll}\\ \textrm{a}.\quad \displaystyle \frac{x^{2}+x+1}{x^{2}-2x-8}> 0&\qquad \textrm{b}.\quad \displaystyle \frac{-x^{2}-x-1}{x^{2}-2x-8}\leq 0 \end{array} \end{array}.

Jawab:

\begin{array}{|l|l|}\hline \multicolumn{2}{|c|}{\textrm{Pertidaksamaan Bentuk Pecahan adalah}\: \: \displaystyle \frac{f(x)}{g(x)}\: \: \textrm{atau}\: \: \displaystyle \frac{h(x)}{k(x)}}\\\hline \displaystyle \frac{x^{2}+x+1}{x^{2}-2x-8}> 0,\quad \textrm{anggap sebagai bentuk}\: \: \frac{f(x)}{g(x)}> 0&\displaystyle \frac{-x^{2}-x-1}{x^{2}-2x-8}\leq 0,\quad \textrm{anggap sebagai bentuk}\: \: \frac{h(x)}{k(x)}\leq 0\\\hline \begin{aligned}f(x)&=x^{2}+x+1\: \: \textrm{adalah definit positif}\\ &\textrm{karena,}\\ &\begin{cases} a=1\: > 0 \\ D=b^{2}-4ac=(1)^{2}-4(1)(1)=-3< 0 \end{cases}\\ &\textrm{sehingga}\\ &\displaystyle \frac{\textcircled{+}}{(x-4)(x+2)}> 0\\ \textrm{HP}&=\left \{ x|x<-2\: \: \textrm{atau}\: \: x> 4,\: \: x\in \mathbb{R} \right \} \end{aligned}&\begin{aligned}h(x)&=-x^{2}-x-1\: \: \textrm{adalah definit negatif}\\ &\textrm{karena}\\ &\begin{cases} a=-1 < 0 \\ D=b^{2}-4ac=(-1)^{2}-4(-1)(-1)=-3< 0 \end{cases}\\ &\textrm{sehingga}\\ &\displaystyle \frac{\textcircled{-}}{(x-4)(x+2)}\leq 0\\ \textrm{HP}&=\left \{ x|x< -2\: \: \textrm{atau}\: \: x> 4,\: \: x\in \mathbb{R} \right \} \end{aligned}\\\hline \end{array}.

Mengapa HP-nya sama?. Kita dapat menggunakan titik uji untuk memastikannya, misalkan sebagai berikut

\begin{array}{|c|c|c|c|}\hline \multicolumn{2}{|l|}{\begin{aligned}f(x)&=x^{2}+x+1\: \: \textrm{adalah definit positif}\\ &\displaystyle \frac{\textcircled{+}}{(x-4)(x+2)}> 0\\ \textrm{HP}&=\left \{ x|x<-2\: \: \textrm{atau}\: \: x> 4,\: \: x\in \mathbb{R} \right \} \end{aligned}}&\multicolumn{2}{|l|}{\begin{aligned}h(x)&=-x^{2}-x-1\: \: \textrm{adalah definit negatif}\\ &\displaystyle \frac{\textcircled{-}}{(x-4)(x+2)}\leq 0\\ \textrm{HP}&=\left \{ x|x< -2\: \: \textrm{atau}\: \: x> 4,\: \: x\in \mathbb{R} \right \} \end{aligned}}\\\hline \multicolumn{4}{|c|}{\textrm{Titik Uji}}\\\hline x=-10&x=10&x=-10&x=10\\\hline \displaystyle \frac{\textcircled{+}}{(-10-4)(-10+2)}=\frac{\textcircled{+}}{\textcircled{+}}=+&\displaystyle \frac{\textcircled{+}}{(10-4)(10+2)}=\frac{\textcircled{+}}{\textcircled{+}}=+&\displaystyle \frac{\textcircled{-}}{(-10-4)(-10+2)}=\frac{\textcircled{-}}{\textcircled{+}}=-&\displaystyle \frac{\textcircled{-}}{(10-4)(10+2)}=\frac{\textcircled{-}}{\textcircled{+}}=-\\\hline \textrm{Benar}&\textrm{Benar}&\textrm{Benar}&\textrm{Benar}\\\hline \end{array}

E.  Pertidaksamaan Bentuk Akar

\textbf{Bentuk Umum}:\: \: \sqrt{f(x)}\cdots \sqrt{g(x)}\: \begin{cases} \sqrt{f(x)}< \sqrt{g(x)}\\\\ \sqrt{f(x)}\leq \sqrt{g(x)} \\\\ \sqrt{f(x)}> \sqrt{g(x)} \\\\ \sqrt{f(x)}\geq \sqrt{g(x)} \end{cases}.

Dalam mengerjakan permasalahan pertidaksamaan bentuk akar (irasional) adalah :

  1. Mengkuadratkan masing-masing ruas.
  2. Di bawah tanda akar (syarat numerus) adalah  \geq .
  3. Himpunan penyelesaian merupakan irisan penyelesaian yang ada.

\LARGE{\fbox{\LARGE{\fbox{Contoh Soal}}}}.

\begin{array}{ll}\\ \fbox{1}.&\textrm{Tentukanlah himpunan penyelesaian dari pertidaksamaan bentuk akar}\\ &\begin{array}{lllllllll}\\ \textrm{a}.\quad \sqrt{x}< 1&\textrm{d}.\quad \sqrt{x}< -1\\ \textrm{b}.\quad \sqrt{x-1}< 2&\textrm{e}.\quad \sqrt{x-1}< -2\\ c.\quad \sqrt{x+1}< 3 &\textrm{f}.\quad \sqrt{x+1}< -3\end{array} \end{array}.

Jawab:

\textbf{a}.\quad \textrm{Untuk}\: \: \sqrt{x}< 1\\ \textrm{(i)}\quad \textrm{Syarat numerus}:x\geq 0\\ \textrm{(ii)}\quad x< 1\qquad \textrm{(dikuadratkan masing-masing ruas)}\\ \textrm{(iii)}\: \: \textrm{HP}=\left \{ x|\: \: 0\leq x< 1,\: x\in \mathbb{R} \right \}\\ \textrm{(iv)}\: \: \textrm{Garis bilangan}.

336

\textbf{b}.\quad \textrm{Untuk}\: \: \sqrt{x-1}< 2\\ \textrm{(i)}\quad \textrm{Syarat numerus}:x-1\geq 0\: \: \Rightarrow \: \: x\geq 1\\ \textrm{(ii)}\quad x-1< 4\qquad \textrm{(dikuadratkan masing-masing ruas)}\: \: \Rightarrow \: \: x< 5\\ \textrm{(iii)}\: \: \textrm{HP}=\left \{ x|\: \: 1\leq x< 5,\: x\in \mathbb{R} \right \}\\ \textrm{(iv)}\: \: \textrm{Garis bilangan}.

337

\textbf{d}.\quad \textrm{Untuk}\: \: \sqrt{x}< -1\\ \textrm{(i)}\quad \textrm{Syarat numerus}:x\geq 0\quad \textrm{(tidak ada bilangan real yang memenuhi)}\\ \textrm{(ii)}\: \: \: \textrm{HP}=\left \{ \right \}.

F.  Pertidaksaan Bentuk Harga Mutlak

Untuk bilangan real x secara definisi nilai mutlak disimbolkan dengan  \left | x \right |.

dengan

\left | x \right |=\begin{cases} x,\quad \textrm{untuk}\: \: x> 0\\ 0,\quad \textrm{untuk}\: \: x=0 \\ -x,\quad \textrm{untuk}\: \: x< 0 \end{cases}.

Beberapa hal yang perlu diperhatikan dalam menyelesaikan pertidaksamaan harga mutlak

\begin{array}{ll}\\ \square &\textrm{untuk}\: a \: \textrm{dan}\: b\: \textrm{bilangan real, berlaku}\\ 1.&\sqrt{a^{2}}=\left | a \right |\: \: \Leftrightarrow \: \: a^{2}=\left | a \right |^{2}\\ 2.&\left | a-b \right |=\begin{cases} a-b ,\: \: \textrm{jika}\: \: a\geq b \\ -(a-b)=b-a\: \: \textrm{jika}\: \: a< b \end{cases}\\ 3.&\left | a-b \right |=\left | b-a \right |\\ 4.&\left | a.b \right |=\left | a \right |.\left | b \right |\\ 5.&\left | \displaystyle \frac{a}{b} \right |=\displaystyle \frac{\left | a \right |}{\left | b \right |},\: \: \textrm{dengan}\: \: b\neq 0\\ 6.&\left | a \right |-\left | b \right |\leq \left | a+b \right |\leq \left | a \right |+\left | b \right |\\ 7.&\left | \displaystyle \frac{a}{b} \right |< c\: \: \Leftrightarrow \: \: \left | a \right |< c\left | b \right |\: \: \Leftrightarrow \: \: (a+cb).(a-cb)< 0\: \: \: (\textrm{di sini} \: \: c> 0)\\ 8.&\left | \displaystyle \frac{a}{b} \right |> c\: \: \Leftrightarrow \: \: \left | a \right |> c\left | b \right |\: \: \Leftrightarrow \: \: (a+cb).(a-cb)> 0\: \: \: (\textrm{di sini} \: \: c> 0)\\ 9.&\left | f(x) \right |< \left | g(x) \right |\: \: \textrm{dapat diubah menjadi}\: \: f^{2}(x)< g^{2}(x)\\ 10.&\left | f(x) \right |> \left | g(x) \right |\: \: \textrm{dapat diubah menjadi}\: \: f^{2}(x)> g^{2}(x) \end{array}.

\LARGE{\fbox{\LARGE{\fbox{Contoh Soal}}}}.

\begin{array}{ll}\\ \fbox{1}.&\textrm{Tentukanlah himpunan penyelesaian yang memenuhi}\\ &\begin{array}{lll}\\ \textrm{a}.\quad \left | 2x-7 \right |=3&\textrm{d}.\quad \left | 2x-7 \right |< 3&\textrm{g}.\quad \left | 2x-7 \right |\leq -3\\ \textrm{b}.\quad \left | 2x-7 \right |=-3&\textrm{e}.\quad \left | 2x-7 \right |\geq 3&\textrm{h}.\quad \left | 2x-7 \right |\geq -3\\ \textrm{c}.\quad \left | x-3 \right |=\left | 2-x \right |&\textrm{f}.\quad \left | x-3 \right |\leq \left | 2-x \right |&\textrm{i}.\quad \left | x^{2}-3x+1 \right |\leq 1 \end{array} \end{array}.

Jawab:

\begin{array}{|l|l|l|}\hline \begin{aligned}\textrm{a}.\quad&\left | 2x-7 \right |=3\\ & 2x-7 =\begin{cases} 3 \\ -3 \end{cases}\\ &2x-7=3\: \: \textrm{atau}\: \: 2x-7=-3\\ &2x=3+7\: \: \textrm{atau}\: \: 2x=-3+7\\ &x=\displaystyle \frac{10}{2}=5\: \: \textrm{atau}\: \: x=\displaystyle \frac{4}{2}=2\\ \textrm{HP}&=\left \{ 2,5 \right \}\\ & \end{aligned}&\begin{aligned}\textrm{b}.\quad&\left | 2x-7 \right |=-3\\ &\textrm{nilai mutlak}\\ &\textrm{suatu bilangan}\\ &\textrm{tidak mungkin}\\ &\textrm{negatif}\\ &\\ \textrm{HP}&=\left \{ \right \}\\ & \end{aligned}&\begin{aligned}\textrm{c}.\quad&\left | x-3 \right |=\left | 2-x \right |\\ &x-3=\begin{cases} (2-x)\\ -(2-x) \end{cases}\\ &x-3=2-x\: \: \textrm{atau}\: \: x-3=-(2-x)\\ &2x=5\: \: \textrm{atau}\: \: -3=-2\: (\textrm{tidak mungkin})\\ &x=\displaystyle \frac{5}{2}\\ \textrm{HP}&=\left \{ \displaystyle \frac{5}{2} \right \} \end{aligned}\\\hline \begin{aligned}\textrm{d}.\quad&\left | 2x-7 \right |< 3\\ &\textrm{maka}\\ &-3< 2x-7< 3\\ &-3+7< 2x-7+7<3+7\\ &4<2x<10\\ &2<x<5\\ \textrm{HP}&=\left \{ x|\: 2<x<5,\: x\in \mathbb{R} \right \}\\ &\\ & \end{aligned}&\begin{aligned}\textrm{e}.\quad&\left | 2x-7 \right |\geq 3\\ &\textrm{maka}\\ &2x-7\leq -3\: \: \textrm{atau}\: \: 2x-7\geq 3\\ &2x\leq -3+7\: \: \textrm{atau}\: \: 2x\geq 3+7\\ &2x\leq 4\: \: \textrm{atau}\: \: 2x\geq 10\\ &x\leq 2\: \: \textrm{atau}\: \: x\geq 5\\ \textrm{HP}&=\left \{ x|\: x\leq 2\: \: \textrm{atau}\: \: x\geq 5,\: x\in \mathbb{R} \right \}\\ &\\ & \end{aligned}&\begin{aligned}\textrm{f}.\quad&\left | x-3 \right |\leq \left | 2-x \right |\\ &\textrm{dikuadratkan kedua ruas}\\ &\left | x-3 \right |^{2}\leq \left | 2-x \right |^{2}\\ &x^{2}-6x+9\leq 4-4x+x^{2}\\ &-6x+4x\leq 4-9\\ &-2x\leq -5\\ &2x\geq 5\\ &x\geq \displaystyle \frac{5}{2}\\ \textrm{HP}&=\left \{ x|\: x\geq \displaystyle \frac{5}{2},\: x\in \mathbb{R} \right \} \end{aligned}\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{2}.&\textrm{Tentukanlah himpunan penyelesaian yang memenuhi}\\ &\begin{array}{lll}\\ \textrm{a}.\quad \left | x-3 \right |=2-x\\ \textrm{b}.\quad \left | x-3 \right |=\left | 2-x \right |\\ \textrm{c}.\quad \left | x-3 \right |\leq \left | 2-x \right |\end{array} \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{array}{|l|l|l|}\hline \begin{aligned}\left | x-3 \right |&=2-x\\ &\textrm{berarti:}\quad &2-x\geq 0\\ &\textrm{sehingga}&x\leq 2\\ \left | x-3 \right |&=\begin{cases} (2-x) \\ -(2-x) \end{cases}\\ x-3&=2-x\: \: \textrm{atau}\: \: x-3=-(2-x)\\ 2x&=5\: \: \textrm{atau}\: \: -3=-2 (\textrm{tak mungkin})\\ x&=\displaystyle \frac{5}{2},\qquad \textrm{padahal}\: \: x\leq 2\\ \textrm{HP}&=\left \{ \right \} \end{aligned}&\begin{aligned}&\textrm{Lihat jawaban/pembahasan}\\ &\textrm{pada No. 1. c} \end{aligned}&\begin{aligned}&\textrm{Lihat jawaban/pembahasan}\\ &\textrm{pada No. 1. f} \end{aligned}\\\hline \end{array}

\begin{array}{ll}\\ \fbox{3}.&\textrm{Tentukanlah himpunan penyelesaian yang memenuhi}\\ &\begin{array}{lll}\\ \textrm{a}.\quad x^{2}-2\left | x \right |-8=0\\ \textrm{b}.\quad x^{2}-2\left | x \right |-8> 0\\ \textrm{c}.\quad x^{2}-2\left | x \right |-8\geq 0\\ \textrm{d}.\quad x^{2}-2\left | x \right |-8< 0\\ \textrm{e}.\quad x^{2}-2\left | x \right |-8\leq 0\end{array} \end{array}.

Jawab:

\begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad&x^{2}-2\left | x \right |-8=0\\ &\left | x \right |^{2}-2\left | x \right |-8=0\\ &\left ( \left | x \right |-4 \right )\left ( \left | x \right |+2 \right )=0\\ &\left | x \right |=4\: \: \textrm{atau}\: \: \left | x \right |=-2\: (\textrm{tidak mungkin})\\ &\left | x \right |=\begin{cases} 4 \\ -4 \end{cases}\\ \textrm{HP}&=\left \{ -4,4 \right \}\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}\textrm{b}.\quad&x^{2}-2\left | x \right |-8> 0\\ &\blacklozenge \: \: (\textrm{untuk}\: \: x\geq 0,\: \: \textrm{maka}\: \: \left | x \right |=x)\\ &x^{2}-2x-8> 0\\ &\left ( x-4 \right )\left ( x+2 \right )> 0\\ \textrm{A}&=\left \{ x|\: x< -2\: \: \textrm{atau}\: \: x> 4,\: x\in \mathbb{R} \right \}\\ &\textrm{karena}\: \: x\geq 0,\: \: \textrm{maka}\\ \textrm{HP}_{1}&=\left \{ x|\: x> 4,\: x\in \mathbb{R} \right \}\\ &\blacklozenge \: \: (\textrm{untuk}\: \: x< 0,\: \: \textrm{maka}\: \: \left | x \right |=-x)\\ &x^{2}+2x-8> 0\\ &\left ( x+4 \right )\left ( x-2 \right )> 0\\ \textrm{B}&=\left \{ x|\: x<-4\: \: \textrm{atau}\: \: x>2,\: x\in \mathbb{R} \right \}\\ &\textrm{karena}\: \: x< 0,\: \: \textrm{maka}\\ \textrm{HP}_{2}&=\left \{ x|\: x< -4 \right \}\\ &\\ \textrm{Jadi}&,\\ \textrm{HP}_{\left (1\: \cup \: 2 \right )}&=\left \{ x|\: x< -4\: \: \textrm{atau}\: \: x> 4,\: x\in \mathbb{R} \right \} \end{aligned}\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{4}.&\textrm{Tentukanlah himpunan penyelesaian yang memenuhi}\\ &\begin{array}{lll}\\ \textrm{a}.\quad \left | 3-\left | x \right | \right |=10\\ \textrm{b}.\quad \left | 3-\left | x \right | \right |> 10\\ \textrm{c}.\quad \left | 3-\left | x \right | \right |\geq 10\\ \textrm{d}.\quad \left | 3-\left | x \right | \right |< 10\\ \textrm{e}.\quad \left | 3-\left | x \right | \right |\leq 10\end{array} \end{array}.

Jawab:

\begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad&\left | 3-\left | x \right | \right |=10\\ &3-\left | x \right |=\begin{cases} 10 \\ -10 \end{cases}\\ &3-\left | x \right |=10\: \: \textrm{atau}\: \: 3-\left | x \right |=-10\\ &\left | x \right |=-7\: \: (\textrm{tidak mungkin})\: \: \textrm{atau}\: \: \left | x \right |=13\\ \textrm{HP}&=\left \{ -13,13 \right \}\\ &\\ &\\ &\end{aligned}&\begin{aligned}\textrm{d}.\quad&\left | 3-\left | x \right | \right |< 10\\ &-10< 3-\left | x \right |< 10\\ &-13< -\left | x \right |< 7\\ &-7< \left | x \right |\leq 13,\: \: (\textrm{ingat harga}\: \: \left | x \right |\geq 0)\\ &0\leq \left | x \right |< 13\\ &\textrm{sehingga},\\ &\left | x \right |< 13\\ &-13< x< 13\\ \textrm{HP}&=\left \{ x|\: -13< x< 13,\: x\in \mathbb{R} \right \} \end{aligned}\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{5}.&\textrm{Himpunan penyelesaian yang memenuhi}\: \: \left | \displaystyle \frac{3-2x}{2+x} \right |\leq 4\\ &\begin{array}{lll}\\ \end{array} \end{array}\\\\ \textrm{Jawab}:\\ \textrm{Perhatikan bahwa soal bentuk di atas memenuhi}\: \: \left | \displaystyle \frac{a}{b} \right |\leq c,\: \: \textrm{dengan}\: \: \: \begin{cases} a&= 3-2x \\ b&=2+x \\ c&= 4 \end{cases}\\ \begin{aligned}\left ( \left (3-2x \right )+4(2+x) \right )\left ( \left (3-2x \right )-4(2+x) \right )&\leq 0\\ (2x+11)(-6x-5)&\leq 0\\ (2x+11)(6x+5)&\geq 0\\ x=-\displaystyle \frac{11}{2}\: \: \textrm{atau}\: \: x=\displaystyle &-\frac{5}{6}\\ \textrm{HP}=\left \{ x|\: x\leq -\displaystyle \frac{11}{2}\: \: \textrm{atau}\: \: x\geq \displaystyle -\frac{5}{6},\: x\in \mathbb{R} \right \}& \end{aligned}.

\begin{array}{ll}\\ \fbox{6}.&\textrm{Nilai-nilai x yang memenuhi}\: \: \left | \displaystyle x+5 \right |>\displaystyle \frac{x-1}{\left | x+1 \right |}\: \: \textrm{adalah ....} \\ &\begin{array}{ll}\\ \textrm{a}.&x<-6\: \: \textrm{atau}\: \: x>1\\ \textrm{b}.&-6<x<-4\\ \textrm{c}.&-\infty < x< -1\: \: \textrm{atau}\: \:-1<x<\infty \\ \textrm{d}.&x>1\\ \textrm{e}.&\textrm{tidak ada nilai x yang memenuhi} \end{array} \\\end{array}.

Jawab: C

Sebagai pengingat x ≠ – 1.

Perhatikan bahwa

\underset{\overbrace{-\infty\qquad\qquad \underset{-x-5\: \: \Leftarrow =\: \: \overbrace{ \left | x+5 \right | }\: \: =\Rightarrow \: \: x+5}{\underbrace{-5}}\qquad\qquad \underset{-x-1\: \: \Leftarrow =\: \: \overbrace{ \left | x+1 \right | }\: \: =\Rightarrow \: \: x+1}{\underbrace{-1}}\qquad\qquad \infty }}{\underbrace{\left | \displaystyle x+5 \right |>\displaystyle \frac{x-1}{\left | x+1 \right |}}}.

\begin{array}{|c|c|c|}\hline \begin{aligned}\textrm{untuk}&\\ &-\infty <x< -5 \end{aligned}&\begin{aligned}\textrm{untuk}&\\ &-5\leq x< -1 \end{aligned}&\begin{aligned}\textrm{untuk}&\\ &-1<x<\infty \end{aligned}\\ \begin{cases} \left | x+5 \right |=-(x+5) \\ \left | x+1 \right |=-(x+1) \end{cases}&\begin{cases} \left | x+5 \right |=(x+5) \\ \left | x+1 \right |=-(x+1) \end{cases}&\begin{cases} \left | x+5 \right |=(x+5) \\ \left | x+1 \right |=(x+1) \end{cases}\\\hline \textbf{Kasus Pertama}&\textbf{Kasus kedua}&\textbf{Kasus ketiga}\\\hline \multicolumn{3}{c}{.}\\ \multicolumn{3}{c}{\textrm{selanjutnya perhatikanlah uraian berikut}} \end{array}.

Kasus Pertama

untuk  x< -5.

\begin{aligned}\left | \displaystyle x+5 \right |&>\displaystyle \frac{x-1}{\left | x+1 \right |}\\ -(x+5)&>\displaystyle \frac{x-1}{-(x+1)}\\ -(x+5)+\displaystyle \frac{x-1}{x+1}&>0 \end{aligned}\quad \begin{aligned}\displaystyle \frac{-(x+5)(x+1)+x-1}{(x+1)}&>0\\ \displaystyle \frac{-x^{2}-5x-6}{(x+1)}&>0\\ \displaystyle \frac{x^{2}+5x+6}{(x+1)}&<0 \end{aligned}\quad \begin{aligned}\displaystyle \frac{(x+2)(x+3)}{(x+1)}&<0\\ \textrm{A}=\left \{ x|\: x<-3\: \textrm{atau}\: -2<x<-1,\: x\in \mathbb{R} \right \}&\\ \textrm{karena}\: \: x<-5&\\ \textrm{HP}_{1}=\left \{ x|\: x<-5 \right \} \end{aligned}.

Kasus kedua

Untuk  -5\leq x< -1.

\begin{aligned}\left | \displaystyle x+5 \right |&>\displaystyle \frac{x-1}{\left | x+1 \right |}\\ (x+5)&>\displaystyle \frac{x-1}{-(x+1)}\\ (x+5)+\displaystyle \frac{x-1}{x+1}&>0 \end{aligned}\quad \begin{aligned}\displaystyle \frac{(x+5)(x+1)+x-1}{(x+1)}&>0\\ \displaystyle \frac{x^{2}-5x-6}{(x+1)}&>0\\ \displaystyle \frac{x^{2}+7x+4}{(x+1)}&>0 \end{aligned}\quad \begin{aligned}\displaystyle \frac{(x+\displaystyle \frac{7+\sqrt{33}}{2})(x+\displaystyle \frac{7-\sqrt{33}}{2})}{(x+1)}&>0\\ \textrm{B}=\left \{ x|\: x<\displaystyle \frac{-7-\sqrt{33}}{2}\: \textrm{atau}\: x>\displaystyle \frac{-7+\sqrt{33}}{2},\: x\in \mathbb{R} \right \}&\\ \textrm{karena}\: \: -5\leq x<-1&\\ \textrm{HP}_{2}=\left \{ x|\: -5\leq x<-1 \right \}& \end{aligned}.

Kasus ketiga

Untuk  x>-1.

\begin{aligned}\left | \displaystyle x+5 \right |&>\displaystyle \frac{x-1}{\left | x+1 \right |}\\ (x+5)&>\displaystyle \frac{x-1}{(x+1)}\\ (x+5)-\displaystyle \frac{x-1}{x+1}&>0 \end{aligned}\quad \begin{aligned}\displaystyle \frac{(x+5)(x+1)-x+1}{(x+1)}&>0\\ \displaystyle \frac{x^{2}+5x+6}{(x+1)}&>0\\ \displaystyle \frac{(x+2)(x+3)}{(x+1)}&>0 \end{aligned}\quad \begin{aligned}\displaystyle \frac{(x+2)(x+3)}{(x+1)}&>0\\ \textrm{C}=\left \{ x|\: -3<x<-2\: \textrm{atau}\: x>-1,\: x\in \mathbb{R} \right \}&\\ \textrm{karena}\: \: x>-1&\\ \textrm{HP}_{3}=\left \{ x|\: x>-1 \right \}& \end{aligned}.

Jadi,  \textrm{HP}=\textrm{HP}_{1}\: \: \cup \: \: \textrm{HP}_{2}\: \: \cup \: \: \textrm{HP}_{3}=\left \{ x|\: x<-1\: \: \textrm{atau}\: \: x>-1,\: x\in \mathbb{R} \right \}.

\begin{array}{ll}\\ \fbox{7}.&\textrm{Jika}\: \: 3<x<5\: \: \textrm{maka penyelesaian untuk} \\ &\sqrt{x^{2}-6x+9}-\sqrt{x^{2}-10x+25}=...\\ &\begin{array}{ll}\\ \textrm{a}.&2x-2\\ \textrm{b}.&2\\ \textrm{c}.&8-2x\\ \textrm{d}.&-2\\ \textrm{e}.&2x-8 \end{array} \\\end{array}\\\\\\ \textrm{Jawab}:\: \textbf{E}\\\\ \begin{aligned}\sqrt{x^{2}-6x+9}-\sqrt{x^{2}-10x+25}&=\sqrt{(x-3)^{2}}-\sqrt{(x-5)^{2}}=\left | x-3 \right |-\left | x-5 \right |\\ \textrm{ingat bahwa saat}\: \: 3<x<5\: \: &\textrm{maka}\: \begin{cases} \left | x-3 \right |=(x-3) \\ \left | x-5 \right |=-(x-5) \end{cases},\: \textrm{sehingga}\\ &=\left | x-3 \right |-\left | x-5 \right |=(x-3)-\left ( -(x-5) \right )\\ &=x-3+x-5\\ &=2x-8 \end{aligned}.

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