Đáp án+Giải thích các bước giải:

Câu `2:` 

`1:)` Thay `x=9(tm)` vào `A`

`=>A=(4sqrt9+1)/(sqrt9-2)=(4*3+1)/(3-2)=13`

Vậy `A=13` tại `x=9`

`2:)`

`B=2/(sqrtx-2)+(3sqrtx)/(sqrtx+2)-(2x)/(x-4)(x>=0;xne4)`

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

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

`B=(2sqrtx+4+3x-6sqrtx-2x)/((sqrtx+2)(sqrtx-2))`

`B=(x-4sqrtx+4)/((sqrtx+2)(sqrtx-2))`

`B=((sqrtx-2)^2)/((sqrtx-2)(sqrtx+2))`

`B=(sqrtx-2)/(sqrtx+2)`

Vậy `B=(sqrtx-2)/(sqrtx+2)` với `x>=0;xne4`

`3:)`

`M=A*B=(4sqrtx+1)/(sqrtx-2)*(sqrtx-2)/(sqrtx+2)=(4sqrtx+1)/(sqrtx+2)=(4(sqrtx+2)-7)/(sqrtx+2)=4-7/(sqrtx+2)`

`x in ZZ` để `M in ZZ`

`=>sqrtx+2 in Ư(7)={+-1;+-7}`

`***` Ta có bảng sau:

\begin{array}{|c|c|c|}\hline \text{$\sqrt{x} +2$}&\text{- 7}&\text{- 1}&\text{1}&\text{7}\\\hline \text{$\sqrt{x}$}&\text{- 9}&\text{- 3}&\text{- 1}&\text{5}\\\hline \text{}&\text{ktm}&\text{ktm}&\text{ktm}&\text{tm}\\\hline \text{x}&\text{}&\text{}&\text{}&\text{25}\\\hline \text{}&\text{}&\text{}&\text{}&\text{tm}\\\hline\end{array}

Vậy `x=25` thì `M in ZZ`

$\color{#FFFF00}{Vi}\color{#CCFF00}{et}\color{#99FF00} {Na}\color{#66FF00}{m}\color{#33FF00}{20}\color{#00FF00} {10}$