Question

Giải phương trình: \(x.\left(x+1\right).\left(x^2+x+1\right)=42\)

Answer 1

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x+1\right)=42\)

Đặt \(x^2+x=t\)

\(\Rightarrow t\left(t+1\right)=42\)

\(\Leftrightarrow t^2+t-42=0\Rightarrow\left[{}\begin{matrix}t=6\\t=-7\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2+x=6\\x^2+x=-7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-6=0\\x^2+x+7=0\left(vô-nghiệm\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)