a) $\frac{3x-2}{3}$ + 5 = $\frac{2(2x+1)}{3}$ + 2x
⇔ 3x - 2 + 3 . 5 = 2( 2x + 1 ) + 3 . 2x
⇔ 3x - 2 + 15 = 4x + 2 + 6x
⇔ 3x - 4x - 6x = 2 - 15 + 2
⇔ -7x = -11
⇔ 7x = 11
⇔ x = $\frac{11}{7}$
b) x³ + 9x² - 4x - 36 = 0
⇔ x²( x + 9 ) - 4( x + 9 ) = 0
⇔ ( x + 9)( x² - 4 ) = 0
⇔ \(\left[ \begin{array}{l}x+9=0\\x²-4=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-9\\x=±2\end{array} \right.\)
~CHÚC BẠN HỌC TỐT~
