4) `x^3-x=0`
⇔`x(x^2-1)=0`
⇔`x(x-1)(x+1)=0`
⇔\(\left[ \begin{array}{l}x=0\\x-1=0\\x+1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\x=1\\x=-1\end{array} \right.\)
Vậy `S={0,1,-1}`
5) `3x(x-10)=x-10`
⇔`3x(x-10)-x+10=0`
⇔`3x(x-10)-(x-10)=0`
⇔`(x-10)(3x-1)=0`
⇔\(\left[ \begin{array}{l}x-10=0\\3x-1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=10\\x=\dfrac{1}{3}\end{array} \right.\)
Vậy `S={10,1/3}`
6) `(x+2)^2-(x+2)=0`
⇔`(x+2)(x+2-1)=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 `S={-2,-1}`
7) `x(x-4)=2x-8`
⇔`x(x-4)-2x+8=0`
⇔`x(x-4)-(2x-8)=0`
⇔`x(x-4)-2(x-4)=0`
⇔`(x-4)(x-2)=0`
⇔\(\left[ \begin{array}{l}x-4=0\\x-2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=4\\x=2\end{array} \right.\)
Vậy `S={4,2}`
8) `x(x-5)=5-x`
⇔`x(x-5)+x-5=0`
⇔`x(x-5)+(x-5)=0`
⇔`(x-5)(x+1)=0`
⇔\(\left[ \begin{array}{l}x-5=0\\x+1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=5\\x=-1\end{array} \right.\)
Vậy `S={5,-1}`
