Question

\(\left(\dfrac{1}{\sqrt[]{x}}+\dfrac{\sqrt[]{x}}{\sqrt[]{x}+1}\right).\dfrac{x+\sqrt[]{x}}{\sqrt[]{x}}\) rút gọn biểu thức

Answer 1

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

 

Answer 2

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

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