Question

\(G=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{x-2\sqrt{3}-3}{x-1}\right)\left(\dfrac{x+3}{x-1}+\dfrac{2}{\sqrt{x}+1}\right)\)

Answer 1

Mình xin phép sửa lại đề nhá.

Ta có \(G=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{x-2\sqrt{x}-3}{x-1}\right)\left(\dfrac{x+3}{x-1}+\dfrac{2}{\sqrt{x}+1}\right)=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\right)\left(\dfrac{x+3+2\sqrt{x}-2}{x-1}\right)=\dfrac{4}{\sqrt{x}-1}.\dfrac{\left(\sqrt{x}+1\right)^2}{x-1}=\dfrac{4}{\sqrt{x}-1}.\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{4\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)