Question

(x^2-x)(x+2)(x+3)=18

Answer 1

\(\Leftrightarrow x\left(x-1\right)\left(x+2\right)\left(x+3\right)-18=0\)

\(\Leftrightarrow x\left(x+2\right)\left(x-1\right)\left(x+3\right)-18=0\)

\(\Leftrightarrow\left(x^2+2x\right)\left(x^2+2x-3\right)-18=0\)

Đặt \(x^2+2x=t\)

\(\Rightarrow t^2-3t-18=0\Rightarrow\left[{}\begin{matrix}t=6\\t=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2+2x=6\\x^2+2x=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+2x-6=0\\x^2+2x+3=0\end{matrix}\right.\) (casio)