a)
\(\sqrt{\left(2x-3\right)^2}=17\)
<=> \(\left|2x-3\right|=17\)
TH1
2x-3=17
<=> 2x=20
<=> x=10
TH2
3-2x=17
<=>-2x=15
<=> x=-15/2
b)\(\sqrt{\left(x+1\right)^2}=2\)
<=>\(\left|x+1\right|=2\)
TH1
x+1=2
x=1
TH2
-x-1=2
-x=3
x=-3
\(2,\\ a,ĐK:x\in R\\ PT\Leftrightarrow\sqrt{\left(2x-3\right)^2}=7\\ \Leftrightarrow\left|2x-3\right|=7\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=7\left(x\ge\dfrac{3}{2}\right)\\2x-3=-7\left(x< \dfrac{3}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=-2\left(tm\right)\end{matrix}\right.\)
\(b,ĐK:x\in R\\ PT\Leftrightarrow\left|x+1\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x+1=2\left(x\ge-1\right)\\x+1=-2\left(x< -1\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-3\left(tm\right)\end{matrix}\right.\\ c,ĐK:x\in R\\ PT\Leftrightarrow\left|x-2\right|=3\\ \Leftrightarrow\left[{}\begin{matrix}x-2=3\left(x\ge2\right)\\2-x=3\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\\ d,ĐK:x\in R\\ PT\Leftrightarrow\left|2x+1\right|=1\\ \Leftrightarrow\left[{}\begin{matrix}2x+1=1\left(x\ge\dfrac{1}{2}\right)\\2x+1=-1\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\Leftrightarrow x=-1\)
\(e,ĐK:x\in R\\ PT\Leftrightarrow\left|3x-1\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}3x-1=5\left(x\ge\dfrac{1}{3}\right)\\3x-1=-5\left(x< \dfrac{1}{3}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-\dfrac{4}{3}\left(tm\right)\end{matrix}\right.\\ f,ĐK:x\ge3\\ PT\Leftrightarrow x-3=49\Leftrightarrow x=52\left(tm\right)\\ g,ĐK:x\ge-\dfrac{5}{2}\\ PT\Leftrightarrow2x+5=16\Leftrightarrow x=\dfrac{11}{2}\left(tm\right)\\ h,ĐK:x\ge\dfrac{3}{5}\\ PT\Leftrightarrow5x-3=4\Leftrightarrow x=\dfrac{7}{5}\left(tm\right)\\ i,ĐK:x\le\dfrac{7}{9}\\ PT\Leftrightarrow7-9x=1\Leftrightarrow x=\dfrac{2}{3}\left(tm\right)\)
\(j,ĐK:x\in R\\ PT\Leftrightarrow2\sqrt{x^2+1}-\dfrac{1}{3}\sqrt{x^2+1}=3\\ \Leftrightarrow\dfrac{5}{3}\sqrt{x^2+1}=3\Leftrightarrow\sqrt{x^2+1}=\dfrac{6}{5}\\ \Leftrightarrow x^2+1=\dfrac{36}{25}\Leftrightarrow x^2=\dfrac{11}{25}\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{11}}{5}\\x=-\dfrac{\sqrt{11}}{5}\end{matrix}\right.\)
à mà lần sau đăng ngắn thôi nhé
làm dài muốn chết :vv
