Đáp án+Giải thích các bước giải:
3)
x.(x-3)-2x+6=0
⇔x(x-3)-(2x-6)=0
⇔x(x-3)-2(x-3)=0
⇔(x-3)(x-2)=0
⇔\(\left[ \begin{array}{l}x-3=0\\x-2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=3\\x=2\end{array} \right.\)
Vậy S={3,2}
4)
$(x-2)^{2}$=(x-2)
⇔$(x-2)^{2}$-(x-2)=0
⇔(x-2)(x-2-1)=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}
5)
($x^{2}$+3).(x+1)+x=-1
⇔($x^{2}$+3).(x+1)+(x+1)=0
⇔(x+1)($x^{2}$+3+1)=0
⇔(x+1)($x^{2}$+4)=0
⇔\(\left[ \begin{array}{l}x+1=0\\(x^{2}+4)=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-1\\x=-2\end{array} \right.\)
Vậy S={-1,-2}
6)
$\frac{x}{3}$+ $\frac{x^{2}}{2}$=0
⇒2x+$3x^{2}$=0
⇔x(2+3x)=0
⇔\(\left[ \begin{array}{l}x=0\\2+3x=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\x=\frac{-2}{3}\end{array} \right.\)
Vậy S={0,$\frac{-2}{3}$}
