a) Ta có: \(3x\left(x-2\right)-2\left(2-x\right)=0\)
\(\Leftrightarrow3x\left(x-2\right)+2\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\frac{2}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-\frac{2}{3}\right\}\)
b) Ta có: \(\left(x+2\right)^2-4x^2=0\)
\(\Leftrightarrow\left(x+2-2x\right)\left(x+2+2x\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2-x=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\frac{2}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-\frac{2}{3}\right\}\)
c) Ta có: \(36-\left(x-4\right)^2=0\)
\(\Leftrightarrow\left(6-x+4\right)\left(6+x-4\right)=0\)
\(\Leftrightarrow\left(10-x\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}10-x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{10;-2\right\}\)
