a. $3x(12x-4)-2x(18x+3)=0$
$⇔36x^2-12x-36x^2-6x=0$
$⇔-18x=0$
$⇔x=0$
Vậy phương trình có nghiệm duy nhất là $x=0$
b. $3x^2-6x=0$
$⇔3x(2x-3)=0$
\(⇔\left[ \begin{array}{l}3x=0\\2x-3=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=0\\x=\dfrac{3}{2}\end{array} \right.\)
Vậy $S=\bigg\{0;\dfrac{3}{2}\bigg\}$
c. $x^2-x+\dfrac{1}{4}=0$
$⇔x^2-2.\dfrac{1}{2}x+\bigg(\dfrac{1}{2}\bigg)^2=0$
$⇔\bigg(x-\dfrac{1}{2}\bigg)=0$
$⇔x=\dfrac{1}{2}$
Vậy phương trình có nghiệm duy nhất là $x=\dfrac{1}{2}$
d. $(2x-3)^2-(x+5)^2=0$
$⇔4x^2-12x+9-x^2-10x-25=0$
$⇔3x^2-22x-16=0$
$⇔3x^2-24x+2x-16=0$
$⇔3x(x-8)+2(x-8)=0$
$⇔(x-8)(3x+2)=0$
\(⇔\left[ \begin{array}{l}x-8=0\\3x+2=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=8\\x=\dfrac{-2}{3}\end{array} \right.\)
Vậy $S=\bigg\{\dfrac{-2}{3};8\bigg\}$
