Question

Cho A= \(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)

a) Tìm đkxđ và rút gọn

b) CM: A>0 với mọi x thuộc đkxđ

c) Tìm GTLN của A

Answer 1

a) điều kiện xác định : \(x\ge0;x\ne1\)

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

\(\Leftrightarrow A=\left(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}-1}{2}\)

\(\Leftrightarrow A=\left(\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}-1}{2}\) \(\Leftrightarrow A=\left(\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\dfrac{2}{\sqrt{x}-1}\) \(\Leftrightarrow A=\dfrac{2}{x+\sqrt{x}+1}\)

a) ta có : \(A=\dfrac{2}{x+\sqrt{x}+1}=\dfrac{2}{\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{1}{4}}>0\forall x\)

c) ta có : \(A=\dfrac{2}{x+\sqrt{x}+1}\le\dfrac{2}{1}=2\) (vì \(x\ge0\) )

\(\Rightarrow\) \(A_{max}=2\) khi \(x=0\)

Answer 2

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