a)\(\left(\left|2-3x\right|+\left|x+1\right|\right)^2=9\)
\(\Leftrightarrow\left(2-3x\right)^2+\left(x+1\right)^2+2\left|\left(2-3x\right)\left(x+1\right)\right|=9\)
\(\Leftrightarrow10x^2-10x+5+2\left|-3x^2-x+2\right|=9\)
\(\Leftrightarrow\left|-3x^2-x+2\right|=\dfrac{4+10x-10x^2}{2}=2+5x-5x^2\)
\(\Leftrightarrow\left[{}\begin{matrix}-3x^2-x+2=2+5x-5x^2\\-3x^2-x+2=-2-5x+5x^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2-6x=0\\8x^2-4x^2-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)
vậy \(S=\left\{-\dfrac{1}{2};0;1;3\right\}\)
b) Trường hợp 1: x+6>0 và x-5>0 suy ra x>5
\(\Leftrightarrow x+6+x-5=3\)
\(\Leftrightarrow x=1\)
Trường hợp 2 x+6<0 và x-5>0 ( vô lí)
Trường hợp 3: x+6>0 và x-5<0 \(\Leftrightarrow-6< 0< 5\)
\(\Leftrightarrow x+6-\left(x-5\right)=3\)
\(11=3\) ( vô lí)
Trường hợp 4 x+6<0 và x-5<0 suy ra x<-6
\(\Leftrightarrow-\left(x+6\right)-\left(x-5\right)=3\)
\(\Leftrightarrow x=-2\)
vậy \(S=\left\{-2;1\right\}\)
a) Vì \(\left|2-3x\right|\) và \(\left|x+1\right|\) luông\(\ge\) 0 với mọi x, y \(\in\) Z.
Mà \(\left|2-3x\right|\)+\(\left|x+1\right|\) =0
\(\Rightarrow\left\{{}\begin{matrix}2-3x=0\\x+1=0\end{matrix}\right.\) =>\(\left\{{}\begin{matrix}3x=2-0\\x=0-1\end{matrix}\right.\) =>\(\left\{{}\begin{matrix}3x=2\\x=-1\end{matrix}\right.\) =>\(\left\{{}\begin{matrix}x=2:3\\x=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\x=-1\end{matrix}\right.\)
Vậy x\(\in\)\(\left\{\dfrac{2}{3};-1\right\}\)
Tick cho mình nha!!!
