Question

Giải phương trình sau:

a) \(\left(x^2+x-2\right)^2+2x^2+2x-4=0\)

Answer 1

Ta có: \(\left(x^2+x-2\right)^2+2x^2+2x-4=0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\x^2+x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\\left(x-1\right)\left(x+2\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=1\\x=-2\end{matrix}\right.\)

Vậy...

Answer 2

 Bạn đặt ẩn phụ \(t=x^2+x-2\left(t\ge-\dfrac{9}{4}\right)\) thì pt thành \(t^2+2t=0\Leftrightarrow\left[{}\begin{matrix}t=0\\t=-2\end{matrix}\right.\) (nhận cả 2 nghiệm)

 Nếu \(t=0\Leftrightarrow x^2+x-2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

 Nếu \(t=-2\Leftrightarrow x^2+x-2=-2\Leftrightarrow x^2+x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

Vậy pt đã cho có tập nghiệm \(S=\left\{-2;-1;0;1\right\}\)