\(\left(2-x\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)=4\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)=-4\)
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)\left(x-1\right)\left(x+1\right)=-4\)
\(\Leftrightarrow\left(x^2-4\right)\left(x^2-1\right)=-4\)
Đặt \(x^2-4=u\)
Phương trình trở thành \(u\left(u+3\right)=-4\)
\(\Leftrightarrow u^2+3u+4=0\)
Mà \(u^2+3u+4=\left(i^2+3u+\frac{9}{4}\right)+\frac{7}{4}=\left(u+\frac{3}{2}\right)^2+\frac{7}{4}>0\)nên phương trình vô nghiệm
