Đặt \(x^2+x+1=a\)
PT <=> a(a+1) = 12
<=> \(a^2+a=12\)
<=> \(\left(a+\frac{1}{2}\right)^2=\frac{49}{4}\)
<=> \(\left[{}\begin{matrix}a=3\\a=-4\end{matrix}\right.\)
TH1: a = 3
<=> \(x^2+x+1=3\)
<=> \(\left(x+\frac{1}{2}\right)^2=\frac{9}{4}\)
<=> \(\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
TH2: a = -4
<=> \(x^2+x+1=-4\)
<=> \(\left(x+\frac{1}{2}\right)^2=\frac{-19}{4}\)
<=> Vô nghiệm
KL: x = {1;-2}
Đặt \(a=x^2+x+1\)
Ta có: \(\left(x^2+x+1\right)\left(x^2+x+2\right)=12\)
\(\Leftrightarrow a\left(a+1\right)-12=0\)
\(\Leftrightarrow a^2+a-12=0\)
\(\Leftrightarrow a^2+4a-3a-12=0\)
\(\Leftrightarrow a\left(a+4\right)-3\left(a+4\right)=0\)
\(\Leftrightarrow\left(a+4\right)\left(a-3\right)=0\)
\(\Leftrightarrow\left(x^2+x+1+4\right)\left(x^2+x+1-3\right)=0\)
\(\Leftrightarrow\left(x^2+x+5\right)\left(x^2+x-2\right)=0\)(1)
Ta có: \(x^2+x+5=x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{19}{4}=\left(x+\frac{1}{2}\right)^2+\frac{19}{4}\ge\frac{19}{4}>0\forall x\)(2)
Từ (1) và (2) suy ra \(x^2+x-2=0\)
\(\Leftrightarrow x^2+2x-x-2=0\)
\(\Leftrightarrow x\left(x+2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
Vậy: \(x\in\left\{-2;1\right\}\)
Đặt \(x^2+x+1=t\)
\(t\left(t+1\right)=12\Leftrightarrow t^2+t-12=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2+x+1=3\\x^2+x+1=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\x^2+x+5=0\left(vn\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
