Question

giúp mình câu  3 ạ

 

Answer 1

1. thay x=9 vào A ta có:

\(A=\dfrac{1-\sqrt{9}}{1+\sqrt{9}}=-\dfrac{1}{2}\)

2. \(\left(\dfrac{6-\sqrt{x}}{x-4}+\dfrac{2}{\sqrt{x}+2}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-2}=\left(\dfrac{6-\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-2}=\dfrac{6-\sqrt{x}+2\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}=\dfrac{1}{\sqrt{x}+1}\)