Question

[Kiệt Nguyễn]

MỌI NGƯỜI Ơ GIÚP MÌNH VỚI MÌNH ĐANG CẦN GẤP

Answer 1

[Nguyễn Việt Lâm]

a.

Khi \(x=9\Rightarrow A=\dfrac{2\sqrt{9}}{\sqrt{9}+2}=\dfrac{6}{5}\)

b.

\(P=\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{3\sqrt{x}}{\sqrt{x}-2}-\dfrac{5x+4}{x-4}\)

\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{3\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{5x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{2x-4\sqrt{x}+3x+6\sqrt{x}-5x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

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

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

c.

Do \(x\ge0\Rightarrow\sqrt{x}+2\ge2\)

\(\Rightarrow\dfrac{2}{\sqrt{x}+2}\le\dfrac{2}{2}=1\)

Vậy \(P_{max}=1\) khi \(x=0\)