`\qquad B=1/(2\sqrt{x}-2)-1/(2\sqrt{x}+2)+\sqrt{x}/(1-x)`
`a)` ĐKXĐ: `{(x>=0),(2\sqrt{x}-2\ne0),(2\sqrt{x}+2\ne0 AAx),(1-x\ne0):}<=>{(x>=0),(\sqrt{x}\ne1),(x\ne1):}<=>{(x>=0),(x\ne1):}`
Với `x>=0;x\ne1` thì
`B=1/(2(\sqrt{x}-1))-1/(2(\sqrt{x}+1))-\sqrt{x}/(x-1)`
`B=(\sqrt{x}+1-(\sqrt{x}-1)-2\sqrt{x})/(2(sqrt{x}-1)(sqrt{x}+1))`
`B=(sqrt{x}+1-sqrt{x}+1-2sqrt{x})/(2(sqrt{x}-1)(sqrt{x}+1))`
`B=(-2\sqrt{x}+2)/(2(sqrt{x}-1)(sqrt{x}+1))`
`B=(-2(sqrt{x}-1))/(2(sqrt{x}-1)(sqrt{x}+1))`
`B=-1/(\sqrt{x}+1)`
Vậy `B=-1/(sqrt{x}+1)` với `x>=0;x\ne1`
`b)` Thay `x=3(\text{tmđk})` vào B ta có:
`B=-1/(\sqrt{3}+1)=(-\sqrt{3}+1)/(3-1)=(-sqrt{3}+1)/2`
Vậy `B=(-sqrt{3}+1)/2` khi `x=3`
