a, Ta có:
(x+2)(3-4x) = x² + 4x + 4
⇔ (x+2)(3-4x) = (x+2)²
⇔ (x+2)( 3-4x-x-2 ) = 0
⇔ (x+2)( 1-5x ) = 0
⇔ \(\left[ \begin{array}{l}x+2=0\\1-5x=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-2\\x=\frac{1}{5}\end{array} \right.\)
b, Ta có:
x³ - 3x + 2 = 0
⇔ (x³ + 2x²) - (2x² + 4x) + (x + 2) = 0
⇔ x²(x+2) - 2x(x+2) + (x+2) = 0
⇔ (x+2)(x²-2x+1) = 0
⇔ (x+2)(x-1)² = 0
⇔ \(\left[ \begin{array}{l}x+2=0\\(x-1)²=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-2\\x=1\end{array} \right.\)
