\(\left(x-1\right)\left(x+2\right)=\left(x-1\right)\left(x^2-x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)-\left(x-1\right)\left(x^2-x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2-x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-x^2+3x-x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[-x\left(x-3\right)-\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(-x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\\x=-1\end{matrix}\right.\)
Vậy....
(x-1)(x+2)=(x-1)(x2-x-1)
\(\Leftrightarrow\)(x-1)(x+2)-(x-1)(x2-x-1)=0
\(\Leftrightarrow\)(x-1)(x+2-x2+x+1)=0
\(\Leftrightarrow\)(x-1)(-x2+2x+3)=0
\(\Leftrightarrow\)(x-1)(-x2+3x-x+3)=0
\(\Leftrightarrow\)(x-1)(3-x)(x+1)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3-x=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\\x=-1\end{matrix}\right.\)
Vậy x\(\in\left\{1;-1;3\right\}\)
