a) `TXĐ: [0; 1)∪(1; \infty); {x|x≥0; x\ne 1}`
`B=1/(2\sqrtx-2)-1/(2\sqrtx+2)+(\sqrtx)/(1-x)`
`B=(\sqrtx+1)/(2x-2)-(\sqrtx-1)/(2x-2)+(-2\sqrtx)/(2x-2)`
`B=(2-2\sqrtx)/(2x-2)`
`B=(1-\sqrtx)/(x-1)`
b) `x=3`
`⇒B=(1-\sqrt3)/(3-1)`
`B=(1-\sqrt3)/2`
c) `|B|=1/2`
`⇒B=(1-\sqrtx)/(x-1)=±1/2`
TH1: `(1-\sqrtx)/(x-1)=1/2`
`⇒2-2\sqrtx=x-1`
`⇒x+2\sqrtx-3=0`
`⇒(\sqrtx+1)^2=4`
`⇒\sqrtx+1=±2`
`⇒\sqrtx=1; -3` (loại `-3`)
`⇒x=1` (loại)
TH2: `(1-\sqrtx)/(x-1)=-1/2`
`⇒-2+2\sqrtx=x-1`
`⇒x-2\sqrtx+1=0`
`⇒\sqrtx=1`
`⇒x=1`
Vậy `x=1`
