`***`Lời giải`***`
a)
`A=(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}):(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1})`
ĐKXĐ: `x>0;xne4;xne1`
`=\frac{\sqrt{x}-(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}-1)}:\frac{(\sqrt{x}+1)(\sqrt{x}-1)-(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}-1)}`
`=\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}(\sqrt{x}-1)}:\frac{x-1+4-x}{(\sqrt{x}-2)(\sqrt{x}-1)}`
`=\frac{1}{\sqrt{x}(\sqrt{x}-1)}.\frac{(\sqrt{x}-2)(\sqrt{x}-1)}{3}`
`=\frac{\sqrt{x}-2}{3\sqrt{x}}`
Vậy `A=\frac{\sqrt{x}-2}{3\sqrt{x}}` với `x>0;xne4;xne1`
b)
Ta có: `A=1/4`
`=>\frac{\sqrt{x}-2}{3\sqrt{x}}=1/4`
`<=>\frac{\sqrt{x}-2}{3\sqrt{x}}-1/4=0`
`<=>\frac{4(\sqrt{x}-2)-3\sqrt{x}}{12\sqrt{x}}=0`
`<=>\frac{4\sqrt{x}-8-3\sqrt{x}}{12\sqrt{x}}=0`
`<=>\frac{\sqrt{x}-8}{12\sqrt{x}}=0`
`=>\sqrt{x}-8=0`
`<=>\sqrt{x}=8`
`<=>x=64(N)`
Vậy `x=64` để `A=1/4`
