\(A=\dfrac{x+16}{\sqrt{x}+3}=\dfrac{x-3^2+25}{\sqrt{x}+3}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+25}{\sqrt{x}+3}\)
\(A=\sqrt{x}-3+\dfrac{25}{\sqrt{x}+3}=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6\)
Áp dụng bđt AM-GM cho 2 số dương ta có:
\(A=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6\ge2\sqrt{\left(\sqrt{x}+3\right)\left(\dfrac{25}{\sqrt{x}+3}\right)}-6\)
\(A\ge2.5-6=4\)
Dấu "=" xảy ra khi \(\sqrt{x}+3=\dfrac{25}{\sqrt{x}+3}\Leftrightarrow\left(\sqrt{x}+3\right)^2=25\)
\(\Rightarrow\sqrt{x}+3=5\) (vì \(\sqrt{x}+3\ge3\forall x\ge0\))
=> \(\sqrt{x}=2\Rightarrow x=4\)
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