Đáp án:+Giải thích các bước giải:

 `a.A=(1-sqrtx)/(1+sqrtx)` `(x>=0)`

Ta có `x=6-2sqrt5`

`x=5-2sqrt5+1`

`x=(sqrt5-1)^2`

`to sqrtx = sqrt5-1`

Thay `sqrtx=sqrt5-1` vào ta được 

`A=(1-(sqrt5-1))/(1+sqrt5-1) = (1-sqrt5+1)/(2+sqrt5)=(2-sqrt5)/(sqrt5)=(2sqrt5-5)/5 .`

`b.B=((15-sqrtx)/(x-25) +2/(sqrtx+5)):(sqrtx+1)/(sqrtx-5)`

`Đk :x>=0 ; xne 25.`

`=(15-sqrtx+2(sqrtx-5))/((sqrtx-5)(sqrtx+5)) * (sqrtx-5)/(sqrtx+1)`

`=(15-sqrtx+2sqrtx-10)/((sqrtx+1)(sqrtx+5)`

`=(sqrtx+5)/((sqrtx+1)(sqrtx+5)`

`=1/(sqrtx+1)`

`c.M=B-A`

`M=1/(sqrtx+1) -(1-sqrtx)/(1+sqrtx)`

`M=(1-1+sqrtx)/(sqrtx+1)`

`M=(sqrtx+1-1)/(sqrtx+1)`

`M=1-1/(sqrtx+1)`

Để `M in ZZ` thì `1/(sqrtx+1) in ZZ`

`to sqrtx+1 in Ư_((1))={-1;1}`

`to sqrtx in {-2;0}`

Mà `sqrtx >=0 ` nên `sqrtx=0 to x=0`

Vậy `x=0.`