Đáp án+Giải thích các bước giải:
a, 4(x+ 2) - 3x(2 + x) = 0
⇔ (x + 2).(4 - 3x) = 0
⇒ \(\left[ \begin{array}{l}x=-2\\x=\dfrac{4}{3}\end{array} \right.\)
b, 3x(x + 1) - 6x² = 0
⇔ 3x(x + 1 - 2x) = 0
⇔ 3x( 1 - x) = 0
⇒ \(\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\)
c, 5x(x - 2) - x + 2 = 0
⇔ 5x(x - 2) - (x - 2) = 0
⇔ (x - 2).(5x - 1) = 0
⇒\(\left[ \begin{array}{l}x=2\\x=\dfrac{1}{5}\end{array} \right.\)
d (x - 3)² = x - 3
⇔ (x - 3)² - (x - 3) = 0
⇔ (x - 3).(x - 3 - 1) = 0
⇔ (x - 3).(x - 4) = 0
⇒ \(\left[ \begin{array}{l}x=3\\x=4\end{array} \right.\)
