Đáp án:

Bài 1: $A=\sqrt{x}$

Bài 2: $A=\dfrac{2(\sqrt{x}+1)^2}{x-4}$

Bài 3: $A=\dfrac{1}{\sqrt{x}+1}$

Giải thích các bước giải:

Bài 1:

$ĐKXĐ: x\ge0,\ x\ne4$

Ta có:

$A=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{1}{\sqrt{x}-2}$

$=\left(\dfrac{\sqrt{x}-2}{x-4}+\dfrac{x}{x-4}+\dfrac{\sqrt{x}+2}{x-4}\right)\cdot(\sqrt{x}-2)$

$=\dfrac{x+2\sqrt{x}}{x-4}\cdot(\sqrt{x}-2)$

$=\dfrac{\sqrt{x}(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{x}+2)}\cdot(\sqrt{x}-2)$

$=\sqrt{x}$

Vậy $A=\sqrt{x}$.

Bài 2:

$ĐKXĐ: x\ge0,\ x\ne4$

Ta có:

$A=\left(\dfrac{1}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{x-4}\right):\dfrac{1}{\sqrt{x}+1}$

$=\left(\dfrac{\sqrt{x}+2}{x-4}+\dfrac{\sqrt{x}}{x-4}\right)\cdot(\sqrt{x}+1)$

$=\dfrac{2\sqrt{x}+2}{x-4}\cdot(\sqrt{x}+1)$

$=\dfrac{2(\sqrt{x}+1)^2}{x-4}$

Vậy $A=\dfrac{2(\sqrt{x}+1)^2}{x-4}$.

Bài 3:

$ĐKXĐ: x\ge0,\ x\ne25$

Ta có: $A=\left(\dfrac{15-\sqrt{x}}{x-25}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}$

$=\left(\dfrac{15-\sqrt{x}}{(\sqrt{x}-5)(\sqrt{x}+5)}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}$

$=\left(\dfrac{15-\sqrt{x}+2(\sqrt{x}-5)}{(\sqrt{x}-5)(\sqrt{x}+5)}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}$

$=\dfrac{\sqrt{x}+5}{(\sqrt{x}-5)(\sqrt{x}+5)}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+1}$

$=\dfrac{1}{\sqrt{x}+1}$

Vậy $A=\dfrac{1}{\sqrt{x}+1}$ với $x\ge0,\ x\ne25$.