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`1)` Thay `x=1/4` vào `A`, ta được:

`A=(sqrt(1/4)+5)/(sqrt(1/4)-2)=(1/2+5)/(1/2-2)=(1/2+10/2)/(1/2-4/2)=(11/2)/(-3/2)=11/2 . (-2)/3=-11/3`

Vậy `A=-11/3` tại `x=1/4`

`2)` 

`B=sqrtx/(sqrtx+2)-3/(2-sqrtx)+(3sqrtx-2)/(x-4) \ (xge0,xne4)`

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

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

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

`3) P=A:B`

`P=(sqrtx+5)/(sqrtx-2) : (sqrtx+2)/(sqrtx-2)=(sqrtx+5)/(sqrtx-2) . (sqrtx-2)/(sqrtx+2)=(sqrtx+5)/(sqrtx+2)`

Xét hiệu `P-1:`

`P-1=(sqrtx+5)/(sqrtx+2)-1=(sqrtx+5-(sqrtx+2))/(sqrtx+2)=3/(sqrtx+2)`

Vì `xge0=>sqrtx+2ge0` nên `3/(sqrtx+2)>0=>P>1`

Xét hiệu `P-sqrtP=sqrtP(sqrtP-1)`

Vì `P>1` nên `sqrtP>1=>sqrtP-1>0`

Do đó, `sqrtP(sqrtP-1)>0=>P>sqrtP`

Vậy `P>sqrtP`

$\color{#29B6F6}{\heartsuit Phuongg \ Li nhh \heartsuit}$