`a)`

`x=9/4 => A=(\sqrt{9/4})/(\sqrt{9/4}-3) = (3/2)/(3/2 -3) = (3/2)/((-3)/2) = -1`

Vậy `A=-1` khi `x=9/4`.

`b)`

`B=7/(\sqrt{x}+1) -12/((\sqrt{x}+1)(3-\sqrt{x}))`

`=(7.(\sqrt{x}-3)+12)/((\sqrt{x}+1)(\sqrt{x}-3))`

`=(7\sqrt{x}-21+12)/((\sqrt{x}+1)(\sqrt{x}-3))`

`=(7\sqrt{x}-9)/((\sqrt{x}+1)(\sqrt{x}-3))`

`M=A-B=(\sqrt{x})/(\sqrt{x}-3) - (7\sqrt{x}-9)/((\sqrt{x}+1)(\sqrt{x}-3))`

`=(\sqrt{x}(\sqrt{x}+1)-7\sqrt{x}+9)/((\sqrt{x}+1)(\sqrt{x}-3))`

`=(x+\sqrt{x}-7\sqrt{x}+9)/((\sqrt{x}+1)(\sqrt{x}-3))`

`=(x-6\sqrt{x}+9)/((\sqrt{x}+1)(\sqrt{x}-3))`

`=(\sqrt{x}-3)^2/((\sqrt{x}+1)(\sqrt{x}-3))`

`=(\sqrt{x}-3)/(\sqrt{x}+1)`

Vậy `M=(\sqrt{x}-3)/(\sqrt{x}+1)`.

`c)`

`M^2 <25/4`

`=> ((\sqrt{x}-3)/(\sqrt{x}+1))^2 < 25/4`

`<=> |(\sqrt{x}-3)/(\sqrt{x}+1)| < 5/2`

`<=> -5/2 < (\sqrt{x}-3)/(\sqrt{x}+1) < 5/2`

Trường hợp `1: (\sqrt{x}-3)/(\sqrt{x}+1) > (-5)/2`

`<=> (\sqrt{x}-3)/(\sqrt{x}+1) + 5/2 >0`

`<=> (2.(\sqrt{x}-3)+5.(\sqrt{x}+1))/(2(\sqrt{x}+1)) >0`

`<=> (2\sqrt{x}-6+5\sqrt{x}+5)/(2(\sqrt{x}+1)) >0`

`<=>(7\sqrt{x}-1)/(2(\sqrt{x}+1)) >0`

Ta có: `\sqrt{x} >=0 =>2(\sqrt{x}+1) >0`

`=> 7\sqrt{x}-1 >0 <=> 7\sqrt{x}>1 <=> x> 1/49`

Trường hợp `2: (\sqrt{x}-3)/(\sqrt{x}+1) < 5/2`

`<=> (\sqrt{x}-3)/(\sqrt{x}+1) - 5/2 <0`

`<=> (2.(\sqrt{x}-3)-5(\sqrt{x}+1))/(2(\sqrt{x}+1)) <0`

`<=> (2\sqrt{x}-6-5\sqrt{x}-5)/(2(\sqrt{x}+1)) <0`

`<=>(-3\sqrt{x}-11)/(2(\sqrt{x}+1)) <0`

Ta có: `\sqrt{x} >=0 =>2(\sqrt{x}+1) >0`

`=> -3\sqrt{x}-11 <0 <=> -3\sqrt{x} <11 <=> \sqrt{x} > (-11)/3` (luôn đúng)

Vậy `x> 1/49` thì `M^2 < 25/4`.