Đáp án:
`B=(2\sqrtx-3)/(\sqrtx+3)`
Giải thích các bước giải:
Với `x≥0;x\ne1`
Ta có:
`B=(2\sqrtx)/(\sqrtx-1)-(11(\sqrtx-1)+8)/(x+2\sqrtx-3)`
`\to B=(2\sqrtx)/(\sqrtx-1)-(11\sqrtx-11+8)/(x-\sqrtx+3\sqrtx-3)`
`\to B=(2\sqrtx)/(\sqrtx-1)-(11\sqrtx-3)/(\sqrtx(\sqrtx-1)+3(\sqrtx-1))`
`\to B=(2\sqrtx)/(\sqrtx-1)-(11\sqrtx-3)/((\sqrtx+3)(\sqrtx-1))`
`\to B=(2\sqrtx(\sqrtx+3)-(11\sqrtx-3))/((\sqrtx+3)(\sqrtx-1))`
`\to B=(2x+6\sqrtx-11\sqrtx+3)/((\sqrtx+3)(\sqrtx-1))`
`\to B=(2x-5\sqrtx+3)/((\sqrtx+3)(\sqrtx-1))`
`\to B=(2x-2\sqrtx-3\sqrtx+3)/((\sqrtx+3)(\sqrtx-1))`
`\to B=(2\sqrtx(\sqrtx-1)-3(\sqrtx-1))/((\sqrtx+3)(\sqrtx-1))`
`\to B=((2\sqrtx-3)(\sqrtx-1))/((\sqrtx+3)(\sqrtx-1))`
`\to B=(2\sqrtx-3)/(\sqrtx+3)`
Vậy với `x≥0;x\ne1` thì `B=(2\sqrtx-3)/(\sqrtx+3)`