Mn giải giúp mình với, mình cần gấp ạ

`#khanhly`

`a,`

`x^3-x=0`

`<=>x(x^2-1)=0`

`<=>x(x-1)(x+1)=0`

`th_1 : x=0`

`th_2 : x-1=0`

`<=>x=1`

`th_3 : x+1=0`

`<=>x=-1`

Vậy `x\in{0;1;-1}`

 `b,`

`x^2-4x-12=0`

`<=>x^2-6x+2x-12=0`

`<=>x(x-6)+2(x-6)=0`

`<=>(x+2)(x-6)=0`

`th_1 : x+2=0`

`<=>x=-2`

`th_2 : x-6=0`

`<=>x=6`

Vậy `x\in{-2;6}`

`c,`

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

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

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

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

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

`th_1 : x-2=0`

`<=>x=2`

`th_2 : x+2=0`

`<=>x=-2`

`th_3 : x+1=0`

`<=>x=-1`

Vậy `x\in{2; -2; -1}`

`d,`

`x(x-2)+6=3x`

`<=>x(x-2)+6-3x=0`

`<=>x(x-2)-(3x-6)=0`

`<=>x(x-2)-3(x-2)=0`

`<=>(x-2)(x-3)=0`

`th_1 : x-2=0`

`<=>x=2`

`th_2 : x-3=0`

`<=>x=3`

Vậy `x\in{2 ; 3}`

`e,`

`x^3-2x^2+x^2=4`

`<=>x^3-2x^2+x^2-4=0`

`<=>x^2(x-2)+(x-2)(x+2)=0`

`<=>(x-2)(x^2+x+2)=0`

`th_1 : x-2=0`

`<=>x=2`

`th_2 : x^2+x+2=0`

`<=>x^2+2 . x . 1/2+(1/2)^2+7/4=0`

`<=>(x+1/2)^2=-7/4` `(` vô lí vì `x^2>=0` `AA` `x` `)`

Vậy `x=2`