Ta có: \(\left(x^2+x\right)^2+\left(x^2+x\right)=6\)
\(\Leftrightarrow\left(x^2+x\right)^2+\left(x^2+x\right)-6=0\)
\(\Leftrightarrow\left(x^2+x\right)^2+3\left(x^2+x\right)-2\left(x^2+x\right)-6=0\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x+3\right)-2\left(x^2+x+3\right)=0\)
\(\Leftrightarrow\left(x^2+x+3\right)\left(x^2+x-2\right)=0\)(1)
Ta có: \(x^2+x+3\)
\(=x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{11}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{11}{4}\)
Ta có: \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}>0\forall x\)
hay \(x^2+x+3>0\forall x\)(2)
Từ (1) và (2) suy ra \(x^2+x-2=0\)
hay \(x^2+2x-x-2=0\)
\(\Leftrightarrow x\left(x+2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
Vậy: \(x\in\left\{-2;1\right\}\)
Bài này đặt \(a=x^2+x\) rồi giải như bình thường nhé! Tìm ra a rồi thì cho cái \(x^2+x=a\) rồi tìm x
