Đáp án:
` B=6/(\sqrtx-3)`
Giải thích các bước giải:
Với `x≥0;x\ne1;x\ne9`
Ta có:
`B=(1/(\sqrtx+3)+(2\sqrtx)/(x-9)).(2\sqrtx+6)/(\sqrtx-1)`
`\to B=(1/(\sqrtx+3)+(2\sqrtx)/((\sqrtx-3)(\sqrtx+3))).(2\sqrtx+6)/(\sqrtx-1)`
`\to B=(\sqrtx-3+2\sqrtx)/((\sqrtx-3)(\sqrtx+3)).(2\sqrtx+6)/(\sqrtx-1)`
`\to B=(3\sqrtx-3)/((\sqrtx-3)(\sqrtx+3)).(2\sqrtx+6)/(\sqrtx-1)`
`\to B=(3(\sqrtx-1))/((\sqrtx-3)(\sqrtx+3)).(2(\sqrtx+3))/(\sqrtx-1)`
`\to B=3/((\sqrtx-3)(\sqrtx+3)).2(\sqrtx+3)`
`\to B=3.2/(\sqrtx-3)`
`\to B=6/(\sqrtx-3)`
Vậy với `x≥0;x\ne1;x\ne9` thì ` B=6/(\sqrtx-3)`
