a) `x^3+2x=0`
⇔`x(x^2+2)=0`
⇔\(\left[ \begin{array}{l}x=0\\x^2+2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\x^2=-2 (vô lý)\end{array} \right.\)
Vậy `S={0}`
b) `x(x+1)+2(x+1)=0`
⇔`(x+1)(x+2)=0`
⇔\(\left[ \begin{array}{l}x+1=0\\x+2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-1\\x=-2\end{array} \right.\)
Vậy `S={-1,-2}`
c) `x(x+1)-x-1=0`
⇔`x(x+1)-(x+1)=0`
⇔`(x+1)(x-1)=0`
⇔\(\left[ \begin{array}{l}x+1=0\\x-1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-1\\x=1\end{array} \right.\)
Vậy `S={-1,1}`
d) `x^2+4=4x`
⇔`x^2-4x+4=0`
⇔`(x-2)^2=0`
⇔`x-2=0`
⇔`x=2`
Vậy `S={2}`
e) `x^2-5x=-6`
⇔`x^2-5x+6=0`
⇔`(x^2-2x)-(3x-6)=0`
⇔`x(x-2)-3(x-2)=0`
⇔`(x-2)(x-3)=0`
⇔\(\left[ \begin{array}{l}x-2=0\\x-3=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=2\\x=3\end{array} \right.\)
Vậy `S={2,3}`
