Đáp án:
Giải thích các bước giải:
a) $2x(x-3) - 3(x-3) = 0$
⇔ $(x-3)(2x - 3) = 0$
⇔ \(\left[ \begin{array}{l}x - 3 =0\\2x - 3=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=3\\x=\dfrac{3}{2}\end{array} \right.\)
b) $x²(x-1) + 4(1-x) = 0$
⇔ $x²(x-1) - 4(x-1) = 0$
⇔ $(x-1)(x² - 4) = 0$
⇔ \(\left[ \begin{array}{l}x - 1 =0\\x² - 4=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=1\\x=±2\end{array} \right.\)
c) $2x(x-5) = (x-5)²$
⇔ $2x(x-5) - (x-5)² = 0$
⇔ $(x-5)(x + 5) = 0$
⇔$ x² - 25 = 0$
⇔ $x² = 25$
⇔ $x = ±5$
d) $(2x-1)² = (4-3x)²$
⇔ $(2x-1)² - (4-3x)² = 0$
⇔ $(2x-1 - 4 + 3x)(2x - 1 + 4 - 3x) = 0$
⇔$ (5x - 5)(-x + 3) = 0$
⇔ \(\left[ \begin{array}{l}x=1\\-x + 3=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=1\\x=3\end{array} \right.\)
e) $2x(3-4x) - 5x²(4x-3) = 0$
⇔ $(4x - 3)(-5x² - 2x) = 0$
⇔ $-x(4x - 3)(5x + 2) = 0$
⇔ \(\left[ \begin{array}{l}x=0\\x=\dfrac{3}{4}\\x=\dfrac{-2}{5}\end{array} \right.\)
