Giải phương trình a , (2x + 1)^2 = (x - 1)^2 b , (x + 6) (3x - 1) = 36 - x ^2 c, x^3 + 3x^2 - 4x - 12 = 0 mn giúp e vs ạ .

Đáp án:

 $a$ $)$ $(2x+1)^2= (x-1)^2$

$⇔ (2x+1)^2-(x-1)^2=0$

$⇔ (2x+1-x+1).(2x+1+x-1)=0$

$⇔ (x+2).3x=0$

$⇔$  \(\left[ \begin{array}{l}x+2=0\\3x=0\end{array} \right.\) $⇔$  \(\left[ \begin{array}{l}x=-2\\x=0\end{array} \right.\) 

$\text{Vậy S={0; -2}}$

$b) (x+6).(3x-1)= 36-x^2$

$⇔ (x+6).(3x-1)- (6-x).(x+6)=0$

$⇔ (x+6).(3x-1-6+x)=0$

$⇔ (x+6).(4x-7)=0$

$⇔$ \(\left[ \begin{array}{l}x+6=0\\4x-7=0\end{array} \right.\) $⇔$  \(\left[ \begin{array}{l}x=-6\\4x=7\end{array} \right.\) $⇔$  \(\left[ \begin{array}{l}x=-6\\x=1,75\end{array} \right.\) 

$\text{Vậy S={-6; 1,75}}$

$c) x^3+3x^2-4x-12=0$

$⇔ x².(x+3)-4.(x+3)=0$

$⇔ (x²-4).(x+3)=0$

$⇔$ \(\left[ \begin{array}{l}x^2-4=0\\x+3=0\end{array} \right.\) $⇔$  \(\left[ \begin{array}{l}x^2=4\\x=-3\end{array} \right.\) $⇔$  \(\left[ \begin{array}{l}x=±2\\x=-3\end{array} \right.\) 

$\text{Vậy S={±2;-3}}$