Question

Tìm min A = \(\dfrac{x+\sqrt{x}}{\sqrt{x}-1}\)

Answer 1

ĐK: \(x\ge0\)

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

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

\(=\sqrt{x}-1+\dfrac{2}{\sqrt{x}-1}+3\)

Áp dụng BĐT AM-GM ta có:

\(A=\sqrt{x}-1+\dfrac{2}{\sqrt{x}-1}+3\ge2\sqrt{\left(\sqrt{x}-1\right).\dfrac{2}{\sqrt{x}-1}}+3\)Hay \(A\ge2\sqrt{2}+3\).

Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}-1=\dfrac{2}{\sqrt{x}-1}\Leftrightarrow\left(\sqrt{x}-1\right)^2=2\)

\(\Leftrightarrow x=\left(\sqrt{2}+1\right)^2=3+2\sqrt{2}\)

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