Question

Answer 1

1: Ta có: \(S=\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{x+\sqrt{x}}\)

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

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

Answer 2

2.

\(S=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}=\dfrac{x-2\sqrt{x}+1+3\sqrt{x}}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-1\right)^2+3\sqrt{x}}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}+3\ge3\)

\(S_{min}=3\) khi \(x=1\)