a) \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=8\)
\(\Leftrightarrow\left[x\left(x+3\right)\right].\left[\left(x+1\right)\left(x+2\right)\right]=8\)
\(\Leftrightarrow\left(x^2+3x\right)\left(x^2+3x+2\right)=8\)
Đặt \(t=x^2+3x+1\) thì
\(\left(t-1\right)\left(t+1\right)=8\)
\(\Leftrightarrow t^2-9=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=3\\t=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+3x+1=3\\x^2+3x+1=-3\end{matrix}\right.\)
Đến đây dễ rồi
b) \(\left(x^2+x\right)\left(x^2+x-2\right)=3\)
Đặt \(t=x^2+x-1\) thì
\(\left(t+1\right)\left(t-1\right)=3\)
\(\Leftrightarrow t^2-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-2\end{matrix}\right.\)
Tương tự bài trên nha
