Đáp án: 1) (2x-1)(x-2)=0
⇔ \(\left[ \begin{array}{l}2x-1=0\\x-2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}2x=1\\x=2\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{1}{2}\\x=2\end{array} \right.\)
2) 3x(2x+5)-5(2x+5)=0
⇔ (2x+5)(3x-5) =0
⇔ \(\left[ \begin{array}{l}2x+5=0\\3x-5=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}2x= -5\\3x=5\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{-5}{2} \\x=\frac{5}{3}\end{array} \right.\)
3) (x+2)(3-4x)+(x²+4x+4)=0
⇔ (x+2)(3-4x)+(x+2)² =0
⇔ (x+2)(3-4x+x+2) = 0
⇔ \(\left[ \begin{array}{l}x+2=0\\-3x+5=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x= -2\\x= \frac{5}{3} \end{array} \right.\)
4) (x²-x)-(3x-3)=0
⇔x(x-1)-3(x-1)=0
⇔ (x-1)(x-3) =0
⇔ \(\left[ \begin{array}{l}x-1=0\\x-3=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=1\\x=3\end{array} \right.\)
