Bài làm:
Ta có: \(\left(x+2\right)\left(x-2\right)\left(x^2-10\right)=72\)
\(\Leftrightarrow\left(x^2-4\right)\left(x^2-10\right)=72\)
\(\Leftrightarrow x^4-14x^2+40-72=0\)
\(\Leftrightarrow x^4-14x^2-32=0\)
\(\Leftrightarrow\left(x^4-16x^2\right)+\left(2x^2-32\right)=0\)
\(\Leftrightarrow x^2\left(x^2-16\right)+2\left(x^2-16\right)=0\)
\(\Leftrightarrow\left(x^2+2\right)\left(x^2-16\right)=0\)
Mà \(x^2+2\ge2>0\left(\forall x\right)\)
\(\Rightarrow x^2-16=0\Leftrightarrow\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x+4=0\end{cases}}\Rightarrow x=\pm4\)
( x + 2 )( x - 2 )( x2 - 10 ) = 72
<=> ( x2 - 4 )( x2 - 10 ) = 72
<=> x4 - 14x2 + 40 - 72 = 0
<=> x4 - 14x2 - 32 = 0
Đặt t = x2 ( \(t\ge0\))
Pt <=> t2 - 14t - 32 = 0
<=> t2 + 2t - 16t - 32 = 0
<=> t( t + 2 ) - 16( t + 2 ) = 0
<=> ( t - 16 )( t + 2 ) = 0
<=> \(\orbr{\begin{cases}t-16=0\\t+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}t=16\\t=-2\end{cases}}\)
\(t\ge0\Rightarrow t=16\)
=> x2 = 16
=> \(x=\pm4\)