Đáp án+Giải thích các bước giải:

`3)M=P*Q`

`M=[1/(\sqrt{x}-1) +(\sqrt{x})/(x-1)]:[(\sqrt{x})/(\sqrt{x}-1)+(\sqrt{x})/(x-1)):(2-(2-x)/(\sqrt{x}+1))] `$(đk:x\geqslant 9;x\ne1)$

`M=(1*(\sqrt{x}+1)+\sqrt{x})/((\sqrt{x}-1)(\sqrt{x}+1)) :[((\sqrt{x}*(\sqrt{x}+1)+\sqrt{x})/((\sqrt{x}-1)(\sqrt{x}+1))) :((2*(\sqrt{x}+1)-(2-x))/(\sqrt{x}+1))]`

`M=(\sqrt{x}+1+\sqrt{x})/((\sqrt{x}-1)(\sqrt{x})) :[((x+\sqrt{x}+\sqrt{x})/((\sqrt{x}-1)(\sqrt{x}+1)))``:((2\sqrt{x}+2-2+x)/(\sqrt{x}+1))]`

`M=(2\sqrt{x}+1)/((\sqrt{x}-1)(\sqrt{x}+1)) :((x+2\sqrt{x})/((\sqrt{x}-1)(\sqrt{x}+1)) ``:(x+2\sqrt{x})/(\sqrt{x}+1))`

`M=(2\sqrt{x}+1)/((\sqrt{x}-1)(\sqrt{x}+1))``:((x+2\sqrt{x})/((\sqrt{x}-1)(\sqrt{x}+1)) ``*(\sqrt{x}+1)/(x+2\sqrt{x}))`

`M=(2\sqrt{x}+1)/((\sqrt{x}-1)(\sqrt{x}+1)) ``:1/(\sqrt{x}-1)`

`M=(2\sqrt{x}+1)/((\sqrt{x}-1)(\sqrt{x}+1))``*(\sqrt{x}-1)/1`

`M=(2\sqrt{x}+1)/(\sqrt{x}+1)`

Để `M<3/2`

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

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

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

`<=>(4\sqrt{x}+2-3\sqrt{x}-3)/(2*(\sqrt{x}+1)) <0`

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

Vì: $\sqrt{x} \geqslant 0$

`<=>`$\sqrt{x}+1\geqslant 1$

`<=>`$\sqrt{x}+1>0$

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

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

`->\sqrt{x}-1<0`

`<=>\sqrt{x}<1`

`<=>x<1`

Kết hợp với điều kiện bài cho ta có:

$0\leqslant x <1$

Vậy với $0\leqslant x<1$ thì `m<3/2`

Giải thích các bước giải:

 `1)` Ta có: `P = 1/(sqrtx-1)+(sqrtx)/(x-1)`

`= (sqrtx+1)/[(sqrtx-1)(sqrtx+1)] + (sqrtx)/[(sqrtx-1)(sqrtx+1)]`

`= (2sqrtx+1)/(x-1)`

Tại `x=25` (thỏa mãn điều kiện) thay vào P ta có:

`P = (2sqrt25+1)/(25-1)=11/24`

Vậy khi `x=25` thì `P = 11/24`

`2)` Ta có: `Q = ((sqrtx)/(sqrtx-1)+(sqrtx)/(x-1)):(2-(2-x)/(sqrtx+1))`

`= [ [sqrtx(sqrtx+1)]/[(sqrtx-1)(sqrtx+1)] + (sqrtx)/[(sqrtx-1)(sqrtx+1)] ]: [ [2(sqrtx+1)]/(sqrtx+1) - (2-x)/(sqrtx+1)]`

`= (x+2sqrtx)/[(sqrtx-1)(sqrtx+1)]: (2sqrtx+2-2+x)/(sqrtx+1)`

`= (x+2sqrtx)/[(sqrtx-1)(sqrtx+1)]. (sqrtx+1)/(x+2sqrtx)`

`= 1/(sqrtx-1)`

Lại có: `M = P : Q = (2sqrtx+1)/[(sqrtx-1)(sqrtx+1)]. (sqrtx-1)`

`= (2sqrtx+1)/(sqrtx+1)`

Vậy `M = (2sqrtx+1)/(sqrtx+1)` với `x>=0;x\ne1`

`3)` ĐKXĐ: `x>=0;x\ne1`

Ta có: `M <3/2`

`<=> (2sqrtx+1)/(sqrtx+1)-3/2<0`

`<=> [2(2sqrtx+1)]/[2(sqrtx+1)] - [3(sqrtx+1)]/[2(sqrtx+1)]<0`

`<=> (4sqrtx+2-3sqrtx-3)/[2(sqrtx+1)]<0`

`<=> (sqrtx-1)/[2(sqrtx+1)]<0`

Vì: `x>=0=>sqrtx>=0=>sqrtx+1>0`

`=> 2(sqrtx+1)>0`

nên `sqrtx-1<0`

`<=>sqrtx<1`

`<=>x<1`

Kết hợp ĐKXĐ ta được `0<=x<1` thì `M<3/2`