giúp tớ với mọi người ơiiii

Giải thích các bước giải:

a.Ta có:

$A=\dfrac{\sqrt{x}-4}{x+8\sqrt{x}+16}-\dfrac{\sqrt{x}}{x-16}$

$\to A=\dfrac{\sqrt{x}-4}{(\sqrt{x}+4)^2}-\dfrac{\sqrt{x}}{(\sqrt{x}-4)(\sqrt{x}+4)}$

$\to A=\dfrac{(\sqrt{x}-4)^2-\sqrt{x}(\sqrt{x}+4)}{(\sqrt{x}-4)(\sqrt{x}+4)^2}$

$\to A=\dfrac{-12\sqrt{x}+16}{(\sqrt{x}-4)(\sqrt{x}+4)^2}$

b.Ta có:

$B=\dfrac{\sqrt{x}}{x-10\sqrt{x}+25}-\dfrac1{x-25}$

$\to B=\dfrac{\sqrt{x}}{(\sqrt{x}-5)^2}-\dfrac1{(\sqrt{x}-5)(\sqrt{x}+5)}$

$\to B=\dfrac{\sqrt{x}(\sqrt{x}+5)-(\sqrt{x}-5)}{\sqrt{x}-5)^2(\sqrt{x}+5)}$

$\to B=\dfrac{x+4\sqrt{x}+5}{(\sqrt{x}-5)^2(\sqrt{x}+5)}$

c.Ta có:

$C=\dfrac{\sqrt{x}+3}{x-6\sqrt{x}+9}-\dfrac{\sqrt{x}}{x-9}$

$\to C=\dfrac{\sqrt{x}+3}{(\sqrt{x}-3)^2}-\dfrac{\sqrt{x}}{(\sqrt{x}+3)(\sqrt{x}-3)}$

$\to C=\dfrac{(\sqrt{x}+3)^2-\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}+3)(\sqrt{x}-3)^2}$

$\to C=\dfrac{9\sqrt{x}+9}{(\sqrt{x}+3)(\sqrt{x}-3)^2}$

d.Ta có:

$D=\dfrac{\sqrt{x}-2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}+2}{x-1}$

$\to D=\dfrac{(\sqrt{x}-2)(\sqrt{x}-1)-(\sqrt{x}+2)(\sqrt{x}+1)}{(\sqrt{x}+1)^2(\sqrt{x}-1)}$

$\to D=\dfrac{ -6\sqrt{x}}{(\sqrt{x}+1)^2(\sqrt{x}-1)}$