`B=(2/(sqrtx-2+3/(2sqrtx+1)-(5sqrtx-7)/(2x-3sqrtx-2)):(2sqrtx+3)/(3x-6sqrtx)(x>0,x ne 4)`
`B=((2(2sqrtx+1)+3(sqrtx-2)-5sqrtx+7)/(2x-3sqrtx-2)).(3x-6sqrtx)/(2sqrtx+3)`
`B=((4sqrtx+2+3sqrtx-6-5sqrtx+7)/((2sqrtx+1)(sqrtx-2))).(3sqrtx(sqrtx-2))/(2sqrtx+3)`
`B=((2sqrtx+3)/((2sqrtx+1)(sqrtx-2))).(3sqrtx(sqrtx-2))/(2sqrtx+3)`
`B=((2sqrtx+3)(3sqrtx)(sqrtx-2))/((2sqrtx+1)(2sqrtx+3)(sqrtx-2))`
`B=(3sqrtx)/(2sqrtx+1)`
`b)B in ZZ`
`=>3sqrtx vdots 2sqrtx+1`
`=>6sqrtx vdots 2sqrtx+1`
`=>6sqrtx+3-3 vdots 2sqrtx+1`
`=>3 vdots 2sqrtx+1`
`=>2sqrtx+1 in Ư(3)={1,3}(do\ 2sqrtx+1>=1)`
`=>2sqrtx in {0,2}`
`=>sqrtx in {0,2}`
`=>x in {0,4}`
Mà `x ne 4`
`=>x=0`
Vậy `x=0` thì `B in ZZ`