Question

giải hộ tui với

a, x(x+1)(x+2)(x+3)=8

b, (x² +x)(x² +x -2)=3

Answer 1

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