Giải thích các bước giải:

b.ĐKXĐ: $x\le -\dfrac52$

$\sqrt{1-x}=\sqrt{6-x}-\sqrt{-5-2x}$ 

$\to 1-x=(\sqrt{6-x}-\sqrt{-5-2x})^2$ 

$\to 1-x=6-x-2\sqrt{6-x}\cdot \sqrt{-5-2x}-5-2x$

$\to 1-x=1-3x-2\sqrt{6-x}\cdot \sqrt{-5-2x}$

$\to2x=-2\sqrt{6-x}\cdot \sqrt{-5-2x}$

$\to x=-\sqrt{6-x}\cdot \sqrt{-5-2x}$

$\to x^2=(6-x)(-5-2x)$

$\to x^2=-30-7x+2x^2$

$\to x^2-7x-30=0$

$\to (x+3)(x-10)=0$

Vì $x\le -\dfrac52$

$\to x+3=0\to x=-3$

c.ĐKXĐ: $x\ge 2$

Ta có:

$\sqrt{x^2-4}-\sqrt{x-2}=0$

$\to \sqrt{(x-2)(x+2)}-\sqrt{x-2}=0$

$\to \sqrt{x-2}(\sqrt{x+2}-1)=0$

$\to \sqrt{x-2}=0\to x-2=2\to x=2$

Hoặc $\sqrt{x+2}-1=0\to \sqrt{x+2}=1\to x+2=1\to x=-1$ loại vì $x\ge 2$