`***`Lời giải`***`
a)
`A=\frac{x}{x-4}+\frac{1}{\sqrt{x} -2}+\frac{1}{\sqrt{x} +2}`
ĐKXĐ: `x≥0;xne4`
`=\frac{x+\sqrt{x} +2+\sqrt{x} -2}{x-4}`
`=\frac{x+2\sqrt{x} }{x-4}`
`=\frac{\sqrt{x}(\sqrt{x}+2) }{(\sqrt{x} -2)(\sqrt{x} +2)}`
`=\frac{\sqrt{x} }{\sqrt{x} -2}`
Vậy `A=\frac{\sqrt{x} }{\sqrt{x} -2}` với `x≥0;xne4`
b)
Ta có: `x=25(N)`
`=>A=\frac{\sqrt{25} }{\sqrt{25} -2}=\frac{5 }{5 -2}=5/3`
Vậy `A=5/3` khi `x=25`
c)
Ta có: `A=-1/3`
`=>\frac{\sqrt{x} }{\sqrt{x} -2}=-1/3`
`<=>\frac{\sqrt{x} }{\sqrt{x} -2}+1/3=0`
`<=>\frac{3\sqrt{x}+\sqrt{x} -2 }{3(\sqrt{x} -2)}=0`
`<=>\frac{4\sqrt{x} -2 }{3(\sqrt{x} -2)}=0`
`=>4\sqrt{x} -2=0`
`<=>4\sqrt{x}=2`
`<=>\sqrt{x}=1/2`
`<=>x=1/4(N)`
Vậy `x=1/4` khi `A=-1/3`
