Ta có: \(\left(x-2\right)\left(x-1\right)\left(x+3\right)\left(x+6\right)=12x^2\)
\(\Leftrightarrow\left[\left(x-2\right)\left(x+3\right)\right]\cdot\left[\left(x-1\right)\left(x+6\right)\right]-12x^2=0\)
\(\Leftrightarrow\left(x^2+x-6\right)\left(x^2+5x-6\right)-12x^2=0\)
Đặt \(a=x^2+x-6\)
\(\Leftrightarrow a\left(a+4x\right)-12x^2=0\)
\(\Leftrightarrow a^2+4xa-12x^2=0\)
\(\Leftrightarrow a^2+6xa-2xa-12x^2=0\)
\(\Leftrightarrow a\left(a+6x\right)-2x\left(a+6x\right)=0\)
\(\Leftrightarrow\left(a+6x\right)\left(a-2x\right)=0\)
\(\Leftrightarrow\left(x^2+x-6+6x\right)\left(x^2+x-6-2x\right)=0\)
\(\Leftrightarrow\left(x^2+7x-6\right)\left(x^2-x-6\right)=0\)
\(\Leftrightarrow\left(x^2+7x-6\right)\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+7x-6=0\\x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+2\cdot x\cdot\frac{7}{2}+\frac{49}{4}-\frac{73}{4}=0\\x=3\\x=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+\frac{7}{2}\right)^2=\frac{73}{4}\\x=3\\x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+\frac{7}{2}=\sqrt{\frac{73}{4}}\\x+\frac{7}{2}=-\sqrt{\frac{73}{4}}\\x=3\\x=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\frac{73}{4}}-\frac{7}{2}=\frac{-7+\sqrt{73}}{2}\\x=-\sqrt{\frac{73}{4}}-\frac{7}{2}=-\frac{7+\sqrt{73}}{2}\\x=3\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{-7+\sqrt{73}}{2};-\frac{7+\sqrt{73}}{2};3;-2\right\}\)
PT <=> (x-2)(x+3)(x-1)(x+6) = 12\(x^2\)
<=> \(\left(x^2+x-6\right)\left(x^2+5x-6\right)=12x^2\)
Gọi \(x^2+3x-6=a\)
PT <=> (a-2x)(a+2x) = 12\(x^2\)
<=> \(a^2-4x^2\) = 12\(x^2\)
<=> \(a^2-16x^2=0\)
<=> \(\left(a-4x\right)\left(a+4x\right)=0\)
<=> \(\left[{}\begin{matrix}a=4x\\a=-4x\end{matrix}\right.\)
TH1: a = 4x
<=> \(x^2+3x-6=4x\)
<=> \(\left(x^2-x+\frac{1}{4}\right)=\frac{25}{4}\)
<=> \(\left(x-\frac{1}{2}\right)^2=\frac{25}{4}\)
<=> \(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Tương tự với TH2
