nhanh ạ giải chi tiết

`ĐKXĐ: x > 0 `

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

`=(1/(sqrtx(sqrtx+1)) - sqrtx/(sqrtx(sqrtx +1))) : sqrtx/(sqrtx+1)^2`

`=(1/(sqrtx(sqrtx+1)) - sqrtx/(sqrtx(sqrtx +1))) . (sqrtx+1)^2/sqrtx`

`=(1-sqrtx)/(sqrtx(sqrtx+1)) . (sqrtx+1)^2/sqrtx`

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

`=(1-x)/x`

Để `A> 1/2` thì:

`(1-x)/x > 1/2`

`<=>(1-x)/x - 1/2 > 0`

`<=>(2(1-x))/(2x) - x/(2x) > 0`

`<=>(2-2x)/(2x) - x/(2x) > 0`

`<=>(2-2x-x)/(2x)> 0`

`<=>(2-3x)/(2x)> 0`

`<=>`\(\left[ \begin{array}{l}\begin{cases} 2-3x>0\\2x>0 \end{cases}\\\begin{cases} 2-3x<0\\2x<0 \end{cases}\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}\begin{cases} x<\dfrac{2}{3}\\x>0 \end{cases}\\\begin{cases} x>\dfrac{2}{3}\\x<0 \text{(vô lý)} \end{cases}\end{array} \right.\) 

`<=>` 0 < x < 2/3`

Vậy `0 < x < 2/3`