`a)` ĐKXĐ: `x>=0; xne9`

Thay `x=36` (tm) vào B ta được:

`B=(sqrt(36)-3)/(sqrt(36)+1)=(6-3)/(6+1)=3/7`

Vậy khi `x=36` thì `B=3/7`

`b)`

`A=(2sqrt(x))/(sqrt(x)+3)+(sqrt(x)+1)/(sqrt(x)-3)+(3-11sqrt(x))/(9-x)`

`A=(2sqrt(x))/(sqrt(x)+3)+(sqrt(x)+1)/(sqrt(x)-3)-(3-11sqrt(x))/(x-9)`

`A=[2sqrt(x)(sqrt(x)-3)+(sqrt(x)+1)(sqrt(x)+3)-(3-11sqrt(x))]/[(sqrt(x)-3)(sqrt(x)+3)]`

`A=(2x-6sqrt(x)+x+4sqrt(x)+3-3+11sqrt(x))/[(sqrt(x)-3)(sqrt(x)+3)]`

`A=(3x+9sqrt(x))/[(sqrt(x)-3)(sqrt(x)+3)]`

`A=(3sqrt(x)(sqrt(x)+3))/[(sqrt(x)-3)(sqrt(x)+3)]`

`A=(3sqrt(x))/(sqrt(x)-3)`

`c)`

`P=AB=(3sqrt(x))/(sqrt(x)-3)*(sqrt(x)-3)/(sqrt(x)+1)=(3sqrt(x))/(sqrt(x)+1)`

`P=(3(sqrt(x)+1)-3)/(sqrt(x)+1)=3-3/(sqrt(x)+1)`

Để `P in ZZ` thì `3/(sqrt(x)+1) in ZZ`

`=> sqrt(x)+1 in Ư(3)={1; 3}` (do `sqrt(x)+1>=1`)

TH1: `sqrt(x)+1=1 => sqrt(x)=0 => x=0` (tm)

TH2: `sqrt(x)+1=3 => sqrt(x)=2 => x=4` (tm)

Vậy `x in {0; 4}`

Bài 2:

`a)`

Thay `x=9` (tm) vào A ta được:

`A=(sqrt(9)+3)/(sqrt(9)+1)=(3+3)/(3+1)=6/4=3/2`

Vậy khi `x=9` thì `A=3/2`

`b)`

`B=(sqrt(x))/(sqrt(x)-2)+(2)/(sqrt(x)+2)+(sqrt(x)+10)/(x-4)`

`B=[sqrt(x)(sqrt(x)+2)+2(sqrt(x)-2)+sqrt(x)+10]/[(sqrt(x)-2)(sqrt(x)+2)]`

`B=(x+2sqrt(x)+2sqrt(x)-4+sqrt(x)+10)/[(sqrt(x)-2)(sqrt(x)+2)]`

`B=(x+5sqrt(x)+6)/[(sqrt(x)-2)(sqrt(x)+2)]`

`B=[(sqrt(x)+2)(sqrt(x)+3)]/[(sqrt(x)-2)(sqrt(x)+2)]`

`B=(sqrt(x)+3)/(sqrt(x)-2)`

`c)`

`P=(A)/(B)=(sqrt(x)+3)/(sqrt(x)+1):(sqrt(x)+3)/(sqrt(x)-2)`

`P=(sqrt(x)+3)/(sqrt(x)+1)*(sqrt(x)-2)/(sqrt(x)+3)`

`P=(sqrt(x)-2)/(sqrt(x)+1)=(sqrt(x)+1-3)/(sqrt(x)+1)=1-3/(sqrt(x)+1)`

Ta có: `sqrt(x)>=0AAx => sqrt(x)+1>=1`

`=> 0<3/(sqrt(x)+1)<=3`

`=> -3<= -3/(sqrt(x)+1)<0`

`=> -2<=1-3/(sqrt(x)+1)<1`

`=> -2<=P<1`

Vì `P in ZZ` nên `P in {-2; -1; 0}`

TH1: `P= -2`

`1-3/(sqrt(x)+1)= -2`

`3/(sqrt(x)+1)=3`

`sqrt(x)+1=1`

`sqrt(x)=0`

`x=0` (tm)

TH2: `P= -1`

`1-3/(sqrt(x)+1)= -1`

`3/(sqrt(x)+1)=2`

`sqrt(x)+1=3/2`

`sqrt(x)=1/2`

`x=1/4` (tm)

Th3: `P=0`

`1-3/(sqrt(x)+1)=0`

`3/(sqrt(x)+1)=1`

`sqrt(x)+1=3`

`sqrt(x)=2`

`x=4` (ktm)

Vậy `x in {0; 1/4}`