a) \(\left(x^2-3x\right)\left(x^2+7x+10\right)=216\Rightarrow x\left(x-3\right)\left(x+2\right)\left(x+5\right)=216\)
\(\Rightarrow x\left(x+2\right)\left(x-3\right)\left(x+5\right)=216\Rightarrow\left(x^2+2x\right)\left(x^2+2x-15\right)=216\)
Đặt \(t=x^2+2x\Rightarrow\) pt trở thành \(t\left(t-15\right)=216\Rightarrow t^2-15t-216=0\)
\(\Rightarrow\left(t+9\right)\left(t-24\right)=0\Rightarrow\left[{}\begin{matrix}t=-9\\t=24\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x^2+2x=-9\\x^2+2x=24\end{matrix}\right.\)
\(TH_1:x^2+2x=-9\Rightarrow x^2+2x+9=0\Rightarrow\left(x+1\right)^2+8=0\) (vô lý)
\(TH_2:x^2+2x=24\Rightarrow x^2+2x-24=0\Rightarrow\left(x-4\right)\left(x+6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-6\end{matrix}\right.\)
b) \(\left(2x^2-7x+3\right)\left(2x^2+x-3\right)+9=0\)
\(\Rightarrow\left(x-3\right)\left(2x-1\right)\left(x-1\right)\left(2x+3\right)+9=0\)
\(\Rightarrow\left(x-3\right)\left(2x+3\right)\left(x-1\right)\left(2x-1\right)+9=0\)
\(\Rightarrow\left(2x^2-3x-9\right)\left(2x^2-3x+1\right)+9=0\)
Đặt \(t=2x^2-3x-9\Rightarrow\) pt trở thành \(t\left(t+10\right)+9=0\)
\(\Rightarrow t^2+10t+9=0\Rightarrow\left(t+1\right)\left(t+9\right)=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=-9\end{matrix}\right.\)
\(TH_1:t=-1\Rightarrow2x^2-3x-9=-1\Rightarrow2x^2-3x-8=0\)
\(\Delta=\left(-3\right)^2-4\left(-8\right).2=73\Rightarrow\left[{}\begin{matrix}x=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{3-\sqrt{73}}{4}\\x=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{3+\sqrt{73}}{4}\end{matrix}\right.\)
\(TH_2:t=-9\Rightarrow2x^2-3x-9=-9\Rightarrow2x^2-3x=0\Rightarrow x\left(2x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)
