Tìm x (1-2x)(1+2x)-x(x+2)(x-2)=0

`(1-2x)(1+2x)-x(x-2)(x+2)=0`$\\$
`<=>1^2-(2x)^2-x(x^2-2^2)=0`$\\$
`<=>(1-4x^2)-(x^3-4x)=0`$\\$
`<=>(1-4x^2)-(x^3-4x)=0`$\\$
`<=>1-4x^2-x^3+4x=0`$\\$
`<=>(1+x^3)-(4x^2+4x)=0`$\\$
`<=>-(x^3-1)-(4x^2+4x)=0`$\\$
`<=>-(x-1)(x^2+x+1+4x)=0`$\\$
`<=>-(x-1)(x^2+5x+1)=0`$\\$
`<=>-(x-1)(x^2+5x+1)=0`$\\$
$\textit{Chia ra làm 2 trường hợp}$ $\\$
`TH_(1)`$\\$
`-(x-1)=0`$\\$
`<=>-x+1=0`$\\$
`<=>x=1(n)`$\\$
`TH_(2)`$\\$
`x^2+5x+1 =0`$\\$
`<=>x^2+5x=-1(l)`

$\longrightarrow$`X=1`

Đáp án+Giải thích các bước giải:

`(1-2x)(1+2x)-x(x+2)(x-2)=0`

`=> [12+(2x)^2]-x(x^2 -2^2)=0`

`=> (1+4.2)-x(x^2 -4)=0`

`=> (1+4.2)-(x^3 -4x)=0`

`=> 1+4.2-x^3 +4x=0`

`=> (1+x^3 )-(4.2-4x)=0` 

`=> -(x^3 -1)-(4.2+4x)=0`

`=>-(x-1)(x^2 +x+1)-4x(x-1)=0`

`=> -(x-1)(x^2 +x+1+4x)=0`

`=> -(x-1)(x^2 +5x+1)=0`

`TH1:``-x+1=0`

`=>-x=-1`

`=>x=1`  `(tm)`

`TH2:``x^2 +5x+1=0`  `(kotm)`

$\begin{array}{|c|}\hline \text{}&\text{Vậy x=1}&\text{}\\\hline\end{array}$