Bài 2: Tìm x
a) Ta có: \(4x\left(x-2\right)+x-2=0\)
\(\Leftrightarrow4x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\4x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\frac{1}{4}\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-\frac{1}{4}\right\}\)
b) Ta có: \(\left(3x-1\right)^2-9=0\)
\(\Leftrightarrow\left(3x-1-3\right)\left(3x-1+3\right)=0\)
\(\Leftrightarrow\left(3x-4\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-4=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=4\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{4}{3}\\x=-\frac{2}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{4}{3};-\frac{2}{3}\right\}\)
c) Ta có: \(x^3-8+\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)+\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4+x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+3x+5\right)=0\)
mà \(x^2+3x+5>0\forall x\)
nên x-2=0
hay x=2
Vậy: x=2
