\(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\)
\(\Leftrightarrow\left(x^2-4\right)\left(x^2-10\right)=72\)
\(\Leftrightarrow\left(x^2-4\right)\left(x^2-4-6\right)=72\\\) ( * )
Đặt \(t=x^2-4\)
Khi đó phương trình ( * ) trở thành:
\(t.\left(t-6\right)=72\)
\(\Leftrightarrow t^2-6t=72\)
\(\Leftrightarrow t^2-6t-72=0\)
\(\Leftrightarrow\left(t-12\right)\left(t+6\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}t-12=0\\t+6=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}t=12\\t=-6\end{matrix}\right.\)
* Với \(t=12\)\(\Rightarrow x^2-4=12\Rightarrow x^2=16\Rightarrow x=+-4\)
* Với \(t=-6\Rightarrow x^2-4=-6\Leftrightarrow x^2=-2\) ( vô lý )
Vậy.............................................
\(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\)
\(\Leftrightarrow\left(x^2-4\right)\left(x^2-10\right)=72\)
\(\Leftrightarrow x^4-10x^2-4x^2+40=72\)
\(\Leftrightarrow x^4-14x^2+40=72\)
\(\Leftrightarrow\left(x^2-7\right)^2=81\)
\(\Leftrightarrow x^2-7=9\)
\(\Leftrightarrow x^2=16\)
\(\Leftrightarrow x=\pm\sqrt{16}=\pm4\).
Vậy .........
