Đáp án:
a, 2x(x-3)-(3-x)=0
⇔ 2x(x-3)+(x-3)=0
⇔(x-3)(2x+1)=0
⇔\(\left[ \begin{array}{l}x-3=0\\2x+1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=3\\x=-1/2\end{array} \right.\)
b, 3x(x+5)-6(x+5)=0
⇔ (x+5)(3x-6)=0
⇔\(\left[ \begin{array}{l}x+5=0\\3x-6=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-5\\x=2\end{array} \right.\)
c, $x^{4}$ - $x^{2}$ =0
⇔ x².(x²-1)=0
⇔ x².(x-1)(x+1)=0
⇔\(\left[ \begin{array}{l}x=0\\x=±1 \end{array} \right.\)
