Bài 3: Tìm x biết
a) $( x + 6 )^{2}$ = 144
⇔ x + 6 = 12
⇒ x = 6
b) x³ + 27 + ( x + 3 )×( x - 9 ) = 0
⇔ x³ + x² - 9x + 3x - 27 + 27 = 0
⇔ x³ + x² - 6x = 0
⇔ x( x² + x - 6 ) = 0
⇔ x( x - 2)×(x + 3) = 0
⇒\(\left[ \begin{array}{l}x=0\\x= 2; x = -3\end{array} \right.\)
c) 2x² - x - 6 = 0
⇔ 2x² - 4x + 3x - 6 = 0
⇔ 2x( x - 2 ) + 3( x - 2 ) = 0
⇔ ( 2x + 3 )×( x - 2 ) = 0
⇒ \(\left[ \begin{array}{l}\frac{-3}{2} \\x= 2\end{array} \right.\)
