Đáp án:

$0<x\le\dfrac{25}{4}$

Giải thích các bước giải:

ĐKXĐ: $x>0; x\ne 25$

$A=\bigg{(}\dfrac{\sqrt x}{\sqrt x-5}+\dfrac{\sqrt x}{\sqrt x+5}\bigg):\dfrac{5\sqrt x}{x-25}$

     $=\sqrt x.\bigg(\dfrac{\sqrt x+5}{x-25}+\dfrac{\sqrt x-5}{x-25}\bigg).\dfrac{x-25}{5\sqrt x}$

     $=\dfrac{\sqrt x+5+\sqrt x-5}{5}$

     $=\dfrac{2\sqrt x}{5}$

$A\le 1$

$⇒\dfrac{2\sqrt x}{5}\le 1$

$⇒2\sqrt x\le 5$

$⇒\sqrt x\le\dfrac{5}{2}$

$⇒0<x\le\dfrac{25}{4}$

Vậy $A\le 1$ khi $0<x\le\dfrac{25}{4}$.

`~rai~`

\(ĐKXĐ:\begin{cases}x\ge 0\\\sqrt{x}-5\ne 0\\x-25\ne 0\\5\sqrt{x}\ne 0\end{cases}\\\Leftrightarrow \begin{cases}x>0\\x\ne 25\end{cases}\\\text{Với }x>0;x\ne 25,\text{ta có:}\\A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-5}+\dfrac{\sqrt{x}}{\sqrt{x}+5}\right):\dfrac{5\sqrt{x}}{x-25}\\\quad=\left[\dfrac{\sqrt{x}(\sqrt{x}+5)}{(\sqrt{x}+5)(\sqrt{x}-5)}+\dfrac{\sqrt{x}(\sqrt{x}-5)}{(\sqrt{x}+5)(\sqrt{x}-5)}\right].\dfrac{x-25}{5\sqrt{x}}\\\quad=\dfrac{x+5\sqrt{x}+x-5\sqrt{x}}{x-25}.\dfrac{x-25}{5\sqrt{x}}\\\quad=\dfrac{2x}{x-25}.\dfrac{x-25}{5\sqrt{x}}\\\quad=\dfrac{2\sqrt{x}}{5}\\A\le 1\\\Leftrightarrow \dfrac{2\sqrt{x}}{5}\le 1\\\Leftrightarrow 2\sqrt{x}\le 5\\\Leftrightarrow \sqrt{x}\le \dfrac{5}{2}\\\Leftrightarrow x\le \dfrac{25}{4}\\\text{Kết hợp với ĐKXĐ}\\\Rightarrow 0<x\le\dfrac{25}{4}.\\\text{Vậy với }0<x\le\dfrac{25}{4}\text{ thì }A\le 1.\)