Question

mn giúp em phần tìm gtnn vs ạ, em cần gấp

Answer 1

a) Ta có: \(A=\left(\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{\sqrt{x}}{\sqrt{x}+2}\)

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

\(=\dfrac{x+2\sqrt{x}}{\sqrt{x}-2}\cdot\dfrac{1}{\sqrt{x}}\)

\(=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\)

b) Thay \(x=7+4\sqrt{3}\) vào A, ta được:

\(A=\dfrac{2+\sqrt{3}+2}{2+\sqrt{3}-2}=\dfrac{4+\sqrt{3}}{\sqrt{3}}=\dfrac{4\sqrt{3}+3}{3}\)

c) Ta có: \(M=\dfrac{x+5}{\sqrt{x}-2}:\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\)

\(=\dfrac{x+5}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)

\(=\dfrac{x+5}{\sqrt{x}+2}\)

\(=\sqrt{x}+2+\dfrac{9}{\sqrt{x}+2}-4\)

\(\Leftrightarrow M\ge2\cdot\sqrt{\left(\sqrt{x}+2\right)\cdot\dfrac{9}{\sqrt{x}+2}}-4\)

\(\Leftrightarrow M\ge2\cdot3-4=6-4=2\)

Dấu '=' xảy ra khi \(\sqrt{x}+2=3\)

\(\Leftrightarrow\sqrt{x}=1\)

hay x=1