Đáp án:
` B=(\sqrtx+1)/(\sqrtx-1)`
Giải thích các bước giải:
Với `x>0;x\ne1`
Ta có:
`B=(3x+\sqrt{9x}-3)/(x+\sqrtx-2)-(\sqrtx+1)/(\sqrtx+2)+(\sqrtx-2)/(\sqrtx).(1/(1-\sqrtx)-1)`
`→B=(3x+\sqrt{9x}-3)/((\sqrtx-1)(\sqrtx+2))-(\sqrtx+1)/(\sqrtx+2)+(\sqrtx-2)/(\sqrtx).(1-(1-\sqrtx))/(1-\sqrtx)`
`\to B=(3x+\sqrt{9x}-3)/((\sqrtx-1)(\sqrtx+2))-(\sqrtx+1)/(\sqrtx+2)+(\sqrtx-2)/(\sqrtx).(1-1+\sqrtx)/(1-\sqrtx)`
`\to B=(3x+\sqrt{9x}-3)/((\sqrtx-1)(\sqrtx+2))-(\sqrtx+1)/(\sqrtx+2)+(\sqrtx-2)/(\sqrtx).(\sqrtx)/(1-\sqrtx)`
`\to B=(3x+\sqrt{9x}-3)/((\sqrtx-1)(\sqrtx+2))-(\sqrtx+1)/(\sqrtx+2)+(\sqrtx-2)/(1-\sqrtx)`
`\to B=(3x+\sqrt{9x}-3)/((\sqrtx-1)(\sqrtx+2))-(\sqrtx+1)/(\sqrtx+2)+(2-\sqrtx)/(\sqrtx-1)`
`\to B=(3x+3\sqrtx-3-(\sqrtx+1)(\sqrtx-1)+(2-\sqrtx)(\sqrtx+2))/((\sqrtx-1)(\sqrtx+2))`
`\to B=(3x+3\sqrtx-3-(x-1)+(4-x))/((\sqrtx-1)(\sqrtx+2))`
`\to B=(3x+3\sqrtx-3-x+1+4-x)/((\sqrtx-1)(\sqrtx+2))`
`\to B=((3x-x-x)+3\sqrtx+(-3+1+4))/((\sqrtx-1)(\sqrtx+2))`
`\to B=(x+3\sqrtx+2)/((\sqrtx-1)(\sqrtx+2))`
`\to B=(x+\sqrtx+2\sqrtx+2)/((\sqrtx-1)(\sqrtx+2))`
`\to B=(\sqrtx(\sqrtx+1)+2(\sqrtx+1))/((\sqrtx-1)(\sqrtx+2))`
`\to B=((\sqrtx+1)(\sqrtx+2))/((\sqrtx-1)(\sqrtx+2))`
`\to B=(\sqrtx+1)/(\sqrtx-1)`
Vậy với `x>0;x\ne1` thì ` B=(\sqrtx+1)/(\sqrtx-1)`
