Consider two cases:
+) If \(x\ge2\)then \(x-2\ge0\Rightarrow\left|x-2\right|=x-2\)
Equation becomes: \(\left(x-2\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)=4\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)=4\)
\(\Leftrightarrow\left(x^2-4\right)\left(x^2-1\right)=4\)(1)
Put \(x^2-4=u\)
(1) becomes: \(u\left(u+3\right)=4\)
\(\Leftrightarrow u^2+3u-4=0\)
We have \(\Delta=3^2+4.4=25,\sqrt{\Delta}=5\)
\(\Rightarrow\orbr{\begin{cases}u=\frac{-3+5}{2}=1\\u=\frac{-3-5}{2}=-4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x^2-4=1\\x^2-4=-4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x^2=5\\x^2=0\end{cases}}\Rightarrow x\in\left\{\pm\sqrt{5};0\right\}\)
+) If \(x< 2\)then \(x-2< 0\Rightarrow\left|x-2\right|=2-x\)
Equation becomes: \(\left(2-x\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)=4\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)=-4\)
\(\Leftrightarrow\left(x^2-4\right)\left(x^2-1\right)=-4\)(2)
Put \(x^2-4=v\)
(2) becomes: \(v\left(v+3\right)=-4\)
\(\Leftrightarrow v^2+3v+4=0\)
But \(v^2+3v+4=\left(v+\frac{3}{2}\right)^2+\frac{7}{4}>0\)
So case two has no value
So \(x\in\left\{\pm\sqrt{5};0\right\}\)
