`|x-2|(x-1)(x+1)(x+2) =4`
`+) TH1 : (x-2)(x-1)(x+1)(x+2) =4`
`<=> (x-2)(x+2)(x-1)(x+1) =4`
`<=> (x²-4).(x²-1)=4`
`<=> x^4-5x²=0`
`=> x=√5`
`x=-√5`
`x=0`
`+) TH2: (2-x)(x-1)(x+1)(x+2) =4`
`<=> (2-x)(2+x)(x-1)(x+1) =4`
`<=> (4-x²).(x²-1)=4`
`=> 5x²-x^4=8`
Đặt `x²=t`
`=> -t²+5t-8=0` ( vô nghiệm )
Vậy `x=√5 ; x=-√5 ; x=0`
`@thew`