Giải thích các bước giải:
$a) (2x +3)² -25 = 0$
$⇔ (2x +3 -5).(2x +3 +5) = 0$
$⇔ (2x -2).(2x +8) = 0$
$⇔ \left[ \begin{array}{l}2x -2=0\\2x +8=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=1\\x=-4\end{array} \right.$
Vậy `S = {1; -4}`
$b) 9x² -(x +2)² = 0$
$⇔ (3x -x -2).(3x +x +2) = 0$
$⇔ (2x -2).(4x +2) = 0$
$⇔ \left[ \begin{array}{l}2x -2=0\\4x +2=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=1\\x=\dfrac{-1}{2}\end{array} \right.$
Vậy `S = {1; -1/2}`
$c) (2x -1)² -(x -4)² = 0$
$⇔ (2x -1 -x +4).(2x -1 +x -4) = 0$
$⇔ (x +3).(3x -5) = 0$
$⇔ \left[ \begin{array}{l}x +3=0\\3x -5=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=-3\\x=\dfrac{5}{3}\end{array} \right.$
Vậy `S = {-3; 5/3}`
$d) (2x -1)² -(x -3)² = 0$
$⇔ (2x -1 -x +3).(2x -1 +x -3) = 0$
$⇔ (x +2).(3x -4) = 0$
$⇔ \left[ \begin{array}{l}x +2=0\\3x -4=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=-2\\x=\dfrac{4}{3}\end{array} \right.$
Vậy `S = {-2; 4/3}`
$e) x³ +x² -4x -4 = 0$
$⇔ x.(x +1) -4.(x +1) = 0$
$⇔ (x +1).(x -4) = 0$
$⇔ \left[ \begin{array}{l}x +1=0\\x -4=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=-1\\x=4\end{array} \right.$
Vậy `S = {-1; 4}`
$g) (2x+1)² -(x-4)² =0 $
$⇔ (2x +1 -x +4).(2x +1 +x -4) = 0$
$⇔ (x +5).(3x -3) = 0$
$⇔ \left[ \begin{array}{l}x +5=0\\3x -3=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=-5\\x=1\end{array} \right.$
Vậy `S = {-5; 1}`
$l) x³ -x² -x +1 = 0$
$⇔ x².(x -1) -(x -1) = 0$
$⇔ (x -1).(x² -1) = 0$
$⇔ \left[ \begin{array}{l}x -1=0\\x² -1=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=1\\x=±1\end{array} \right.$
Vậy `S = {±1}`
