- Với \(x\ge2\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)-4=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4\right)-4=0\)
\(\Leftrightarrow x^4-5x^2=0\Leftrightarrow x^2\left(x^2-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=-\sqrt{5}\left(l\right)\\x=\sqrt{5}\end{matrix}\right.\)
- Với \(x< 2\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)+4=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4\right)+4=0\)
\(\Leftrightarrow x^4-5x^2+8=0\left(vn\right)\)
Vậy pt có nghiệm duy nhất \(x=\sqrt{5}\)
