Môn Toán Lớp 8: Giải phương trình m) (x^2 – 16)^2 – (x – 4)^2 = 0 n) (5x^2 – 2x + 10)^2 = (3x^2 + 10x – 8)^2

Giải đáp:

 $m) (x^2-16)^2-(x-4)²=0$

$⇔ (x²-16-x+4).(x²-16+x-4)=0$

$⇔ [(x-4).(x+4)-1.(x-4)].[(x-4).(x+4)+1.(x-4)]=0$

$⇔ (x-4).(x+4-1).(x-4).(x+4+1)=0$

$⇔ (x-4)².(x+3).(x+5)=0$

$⇔$ \(\left[ \begin{array}{l}(x-4)^2=0\\x+3=0\\x+5=0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=4\\x=-3\\x=-5\end{array} \right.\) 

$\text{Vậy S={4;-3;-5}}$

$n) (5x^2-2x+10)^2= (3x^2+10x-8)^2$

$⇔ (5x²-2x+10)^2- (3x^2+10x-8)^2=0$

$⇔ (5x^2-2x+10-3x²-10x+8).(5x²-2x+10+3x²+10x-8)=0$

$⇔ (2x²-12x+18).(8x²+8x+2)=0$

$⇔ [2.(x²-6x+9)].[2.(4x²+4x+1)]=0$

$⇔ 4.(x-3)^2.(2x+1)^2=0$

$⇔$  \(\left[ \begin{array}{l}(x-3)^2=0\\(2x+1)^2=0\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x=3\\x=-0,5\end{array} \right.\) 

$\text{Vậy S={3;-0,5}}$