Question

Gỉai pt :

A = \(\left(x-2\right)\left(x+2\right)+4\left(x-2\right)\sqrt{\frac{x+2}{x-2}}=-3\)

Answer 1

$Dkxd:x>2\text{ hoặc } x\le -2$.

Th1: $x>2$. Khi đó:

$pt\iff (x-2)(x+2)+4\sqrt{x-2}\sqrt{\frac{(x+2)(x-2)}{x-2}}=-3$

$\iff (x-2)(x-2)+4\sqrt{(x-2)(x+2)}+3=0\iff (\sqrt{(x-2)(x+2)}+1)(\sqrt{(x-2)(x+2)}+3)=0(1)$.

Do $\sqrt{(x-2)(x+2)}\ge 0$ nên $VT(1)>0=VP(2)\implies $ vô nghiệm.

Th2: $x\le -2\implies 2-x\ge 0;-x-2>0$.

Khi đó: $pt\iff (2-x)(-x-2)-4(2-x)\sqrt{\frac{-x-2}{2-x}}+3=0$

$\iff (2-x)(-x-2)-4\sqrt{(2-x)(-x-2)}+3=0\iff (\sqrt{(2-x)(-x-2)-1})(\sqrt{(2-x)(-x-2)}-3)=0$.

$\iff \sqrt{(x-2)(x+2)}=1\text{ hoặc } \sqrt{(x-2)(x+2)}=3$.

$\iff x=5(l)\text{ hoặc} x=13(l)$.

Vậy phương trình đã cho vô nghiệm