Môn Toán Lớp 8: Giải các phương trình sau: (4x + 3)(x ² – 9) = (x + 3)(16x ² – 9) (x – 4) ² – 25 = 0 (x – 3) ² – (x + 1) ² = 0 Giúp với

Giải đáp:

$a) (4x+3).(x^2-9)=(x+3).(16x^2-9)$

$⇔ (4x+3).(x-3).(x+3)- (x+3).(4x-3).(4x+3)=0$

$⇔ (4x+3).(x+3).(x-3-4x+3)=0$

$⇔ (4x+3).(x+3).(-3x)=0$

$⇔$ \(\left[ \begin{array}{l}4x+3=0\\x+3=0\\-3x=0\end{array} \right.\)$⇔$  \(\left[ \begin{array}{l}4x=-3\\x=-3\\x=0\end{array} \right.\) $⇔$  \(\left[ \begin{array}{l}x=-0,75\\x=-3\\x=0\end{array} \right.\) 

$\text{Vậy S={-0,75;-3;0}}$

$b) (x-4)^2-25=0$

$⇔ (x-4)^2-5^2=0$

$⇔ (x-4-5).(x-4+5)=0$

$⇔ (x-9).(x+1)=0$

$⇔$\(\left[ \begin{array}{l}x-9=0\\x+1=0\end{array} \right.\) $⇔$  \(\left[ \begin{array}{l}x=9\\x=-1\end{array} \right.\)

$\text{Vậy S = {9;-1}}$

$c) (x-3)^2-(x+1)^2=0$

$⇔ (x-3-x-1).(x-3+x+1)=0$

$⇔ -4.(2x-2)=0$

$⇔ -8x+8=0$

$⇔ -8x=-8$

$⇔ x=1$

$\text{Vậy S={1}}$