a) $(2x+1)(x-1)=0$
\(⇔\left[ \begin{array}{l}2x+1=0\\x-1=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=\dfrac{-1}{2}\\x=1\end{array} \right.\)
Vậy $S=\bigg\{\dfrac{-1}{2};1\bigg\}$
b) $(3x-1)(2x-3)(x+5)=0$
\(⇔\left[ \begin{array}{l}3x-1=0\\2x-3=0\\x+5=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=\dfrac{1}{3}\\x=\dfrac{3}{2}\\x=-5\end{array} \right.\)
Vậy $S=\bigg\{\dfrac{1}{3};\dfrac{3}{2};5\bigg\}$
c) $3x-15=2x(x-5)$
$⇔3(x-5)-2x(x-5)=0$
$⇔(x-5)(3-2x)=0$
\(⇔\left[ \begin{array}{l}x-5=0\\3-2x=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=5\\x=\dfrac{3}{2}\end{array} \right.\)
Vậy $S=\bigg\{5;\dfrac{3}{2}\bigg\}$
d) $x^2-x=0$
$⇔x(x-1)=0$
\(⇔\left[ \begin{array}{l}x=0\\x-1=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\)
Vậy $S=\{0;1\}$
e) $x^2-2x=0$
$⇔x(x-2)=0$
\(⇔\left[ \begin{array}{l}x=0\\x-2=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=0\\x=2\end{array} \right.\)
Vậy $S=\{0;2\}$
f) $x^2-3x=0$
$⇔x(x-3)=0$
\(⇔\left[ \begin{array}{l}x=0\\x-3=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\)
Vậy $S=\{0;3\}$
g) $(x+1)(x+2)=(2-x)(x+2)$
$⇔(x+1)(x+2)-(2-x)(x+2)=0$
$⇔(x+2)(x+1-2+x)=0$
$⇔(x+2)(2x-1)=0$
\(⇔\left[ \begin{array}{l}x+2=0\\2x-1=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=-2\\x=\dfrac{1}{2}\end{array} \right.\)
Vậy $S=\bigg\{-2;\dfrac{1}{2}\bigg\}$
