Question

c \(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{2}{\sqrt{x+2}}\)

d \(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x+3}}\)

Answer 1

hình như đề cứ sao sao á

 

Answer 2

\(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{2}{\sqrt{x}+2}\\ \dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+3}\)

Hay giống trên kia vậy ạ?

Answer 3

\(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{2}{\sqrt{x}+2}\left(\text{đ}k\text{x}\text{đ}:x\ge0;x\ne4\right)\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{x+2\sqrt{x}-2\sqrt{x}+4}{x-4}\\ =\dfrac{x+4}{x-4}\)

\(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+3}\left(\text{đ}k\text{x}\text{đ}:x\ge0;x\ne9\right)\\ =\dfrac{\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{\sqrt{x}+3-\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{6}{x-9}\)