Question

Rút gọn biểu thức

\(\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\right).\dfrac{x-6\sqrt{x}+9}{\sqrt{x}}\)

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

Answer 1

\(a,=\dfrac{\sqrt{x}-3+\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}}\left(x>0;x\ne9\right)\\ =\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\\ b,=\dfrac{x}{\sqrt{x}+1}+\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\left(x>0\right)\\ =\dfrac{x}{\sqrt{x}+1}+\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}=\sqrt{x}+1\)