b/
`x(x-1) -2(1-x) =0`
`<=> x(x-1) + 2(x-1) =0`
`<=> (x+2)(x-1) =0`
`<=>` \(\left[ \begin{array}{l}x+2=0\\x-1=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=-2\\x=1\end{array} \right.\)
Vậy `x \in {-2 ;1}`
c/
`2x(x-2) -(2-x)^2 =0`
`<=> 2x(x-2) - (x-2)^2 =0`
`<=> (x-2)(2x - x +2) =0`
`<=> (x-2)(x+2) =0`
`<=>` \(\left[ \begin{array}{l}x-2=0\\x+2 =0 \end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=2\\x=-2\end{array} \right.\)
Vậy `x \in {2 ; -2}`
d/
`(x-3)^3 +(3-x) =0`
`<=> (x-3)^3 - (x-3) =0`
`<=> (x-3)[(x-3)^2 -1] =0`
`<=>` \(\left[ \begin{array}{l}x-3=0\\(x-3)^2 -1 =0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=3\\(x-3)^2= 1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=3\\x =4\\x =2\end{array} \right.\)
Vậy `x \in {2 ;3 ;4}`
