a) 3(x -1)(x +1) = x+1
⇔ 3(x -1)(x +1) - (x+1) = 0
⇔ (3x - 3)(x+1) - (x+1) = 0
⇔ ( x+1 )( 3x - 3 - 1 ) = 0
⇔ ( x+1 )( 3x - 4 ) = 0
⇔ \(\left[ \begin{array}{l}x+1=0\\3x-4=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-1\\x=4/3\end{array} \right.\)
Vậy S = { -1; $\frac{4}{3}$ }
b) 9x(x -5) = (5 -x)(5+x)
⇔ 9x(x -5) - (5 -x)(5+x) = 0
⇔ 9x(x -5) + (x-5)(5+x) = 0
⇔ (x-5)(9x + 5 + x) = 0
⇔ (x-5)(10x +5 )= 0
⇔ \(\left[ \begin{array}{l}x-5=0\\10x+5=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=5\\x=-1/2\end{array} \right.\)
Vậy S = { 5; $\frac{-1}{2}$ }
