Question

Giups m với hepllllllllll

Answer 1

ĐKXĐ: \(x>0;x\ne1\)

\(\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)\)

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

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

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