Đáp án+Giải thích các bước giải:

 `a)` `5x(x-4)-2x+8=0`

`<=>5x(x-4)-2(x-4)=0`

`<=>(x-4)(5x-2)=0`

`<=>`$\left[\begin{matrix} x-4=0\\ 5x-2=0\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=4\\ x=\dfrac{2}{5}\end{matrix}\right.$

`text{Vậy}` `S={4;2/5}`

 `b)` `x^2-4x+4=2(x-2)`

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

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

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

`<=>(x-2)(x-4)=0`

`<=>`$\left[\begin{matrix} x-2=0\\ x-4=0\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=2\\ x=4\end{matrix}\right.$

`text{Vậy}` `S={2;4}`

 `c)` `(x-1)^3=2x-2`

`<=>(x-1)^3-(2x-2)=0`

`<=>(x-1)^3-2(x-1)=0`

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

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

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

`<=>`$\left[\begin{matrix} x-1=0\\ x-1-\sqrt{2}=0\\ x-1+\sqrt{2}=0\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=1\\ x=1+\sqrt{2}\\ x=1-\sqrt{2}\end{matrix}\right.$

`text{Vậy}` `S={1;1-sqrt(2);1+sqrt(2)}`

a) `5x(x-4)-2x+8=0`

⇔`5x(x-4)-(2x-8)=0`

⇔`5x(x-4)-2(x-4)=0`

⇔`(x-4)(5x-2)=0`

⇔\(\left[ \begin{array}{l}x-4=0\\5x-2=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=4\\x=\dfrac{2}{5}\end{array} \right.\) 

Vậy `S={4,2/5}`

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

⇔`(x-2)^2-2(x-2)=0`

⇔`(x-2)(x-2-2)=0`

⇔`(x-2)(x-4)=0`

⇔\(\left[ \begin{array}{l}x-2=0\\x-4=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=2\\x=4\end{array} \right.\) 

Vậy `S={2,4}`

c) `(x-1)^3=2x-2`

⇔`(x-1)^3-=2(x-1)`

⇔`(x-1)^3-2(x-1)=0`

⇔`(x-1)[(x-1)^2-2]=0`

⇔`(x-1)(x^2-2x+1-2)=0`

⇔`(x-1)(x^2-2x-1)=0`

⇔\(\left[ \begin{array}{l}x-1=0\\x^2-2x-1=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=1\\x=1-\sqrt{2}\\x=1+\sqrt{2}\end{array} \right.\) 

Vậy `S={1,1+\sqrt{2},1-\sqrt{2}}`