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

Đáp án:

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

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

$⇔ (4x+3).(x+3).(x-3-4x+3)=0$

$⇔ (4x+3).(x+3).(-3x)=0$

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

$\text{Vậy S={-0,75;-3;0}}$

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

$⇔ (x-4)^2-5^2=0$

$⇔ (x-4-5).(x-4+5)=0$

$⇔ (x-9).(x+1)=0$

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

$\text{Vậy S = {9;-1}}$

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

$⇔ (x-3-x-1).(x-3+x+1)=0$

$⇔ -4.(2x-2)=0$

$⇔ -8x+8=0$

$⇔ -8x=-8$

$⇔ x=1$

$\text{Vậy S={1}}$ 

$(4x+3)(x^{2}-9)=(x+3)(16x^{2}-9)$

$⇔(4x+3)(x-3)(x+3)=(x+3)(4x-3)(4x+3)$

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

$⇔(x+3)(4x+3)[(x-3)-(4x-3)]=0$

$⇔(x+3)(4x+3)(x-3-4x+3)=0$

$⇔(x+3)(4x+3)(x-4x)=0$

$⇔(x+3)(4x+3)x(1-4)=0$

$⇔(x+3)(4x+3)(-3x)=0$

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

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

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

$Vậy$ $pt$ $có$ $tập$ $nghiệm$ $S=${$-3;\frac{-3}{4};0$}

$(x-4)^{2}-25=0$

$⇔(x-4-5)(x-4+5)=0$

$⇔(x-9)(x+1)=0$

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

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

$Vậy$ $pt$ $trên$ $có$ $tập$ $nghiệm$ $S=${$9;-1$}

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

$⇔(x-3-x-1)(x-3+x+1)=0$

$⇔-4(2x-2)=0$

$⇔2x-2=0$

$⇔2x=2$

$⇔x=1$

$Vậy$ $pt$ $có$ $tập$ $nghiệm$ $S=${$1$}