\(\left(x^2+3x+2\right)\left(x^2+9x+18\right)=168x^2\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)=168x^2\)
\(\Leftrightarrow\left[\left(x+1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]=168x^2\)
\(\Leftrightarrow\left(x^2+7x+6\right)\left(x^2+5x+6\right)=168x^2\)(1)
Đặt \(x^2+5x+6=t\)
Khi đó (1) trở thành: \(\left(t+2x\right)t=168x^2\Leftrightarrow t^2+2xt-168x^2=0\)
\(\Leftrightarrow\left(t-12x\right)\left(t+14x\right)=0\Leftrightarrow\orbr{\begin{cases}t=12x\\t=-14x\end{cases}}\)
TH1: \(t=12x\Rightarrow x^2+5x+6=12x\)
\(\Leftrightarrow x^2-7x+6=0\Leftrightarrow\left(x-1\right)\left(x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=6\end{cases}}\)
TH2: \(t=-14x\Rightarrow x^2+5x+6=-14x\Rightarrow x^2+19x+6=0\)
\(\Leftrightarrow x^2+2.x.\frac{19}{2}+\left(\frac{19}{2}\right)^2-\frac{337}{4}=0\)
\(\Leftrightarrow\left(x+\frac{19}{2}\right)^2=\frac{337}{4}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{337}-19}{2}\\x=\frac{-\sqrt{337}-19}{2}\end{cases}}\)
