Question

Giá trị nhỏ nhất của

A=\(\dfrac{x+8}{\sqrt{x}+1}\)

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Answer 1

\(A=\dfrac{x+8}{\sqrt{x}+1}=\dfrac{4\left(\sqrt{x}+1\right)+x-4\sqrt{x}+4}{\sqrt{x}+1}\)

\(=4+\dfrac{x-4\sqrt{x}+4}{\sqrt{x}+1}=4+\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}\ge4\)

Dấu = xảy ra khi \(x=4\)

Answer 2

Cách khác:

\(A=\dfrac{x+8}{\sqrt{x}+1}=\dfrac{x-1}{\sqrt{x}+1}+\dfrac{9}{\sqrt{x}+1}=\sqrt{x}-1+\dfrac{9}{\sqrt{x}+1}\)

\(=\sqrt{x}+1+\dfrac{9}{\sqrt{x}+1}-2\ge2\sqrt{\left(\sqrt{x}+1\right).\dfrac{9}{\sqrt{x}+1}}-2=4\)

Dấu "=" xảy ra khi x = 4

Answer 3

Cách khác:

\(A=\dfrac{x+8}{\sqrt{x}+1}=\dfrac{x-1}{\sqrt{x}+1}+\dfrac{9}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x+1}\right)}{\sqrt{x+1}}+\dfrac{9}{\sqrt{x}+1}\)

\(=\sqrt{x}-1+\dfrac{9}{\sqrt{x}+1}=\sqrt{x}+1+\dfrac{9}{\sqrt{x}+1}-2\ge2\sqrt{\left(\sqrt{x}+1\right).\dfrac{9}{\sqrt{x}+1}}-2=4\)

Dấu "=" xảy ra khi x = 4