8) Vì \(x>0;x\ne1\Rightarrow x\ge2\Rightarrow\sqrt{x}\ge\sqrt{2}\)
\(A=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}=\dfrac{x+2\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+\dfrac{1}{\sqrt{x}}+2=\dfrac{1}{2}\sqrt{x}+\dfrac{1}{\sqrt{x}}+\dfrac{1}{2}\sqrt{x}+2\)
Áp dụng BĐT cô si cho 2 số dương \(\dfrac{1}{2}\sqrt{x}\) và \(\dfrac{1}{\sqrt{x}}\), ta có:
\(\dfrac{1}{2}\sqrt{x}+\dfrac{1}{\sqrt{x}}\ge2\sqrt{\dfrac{1}{2}\sqrt{x}.\dfrac{1}{\sqrt{x}}}=2.\dfrac{1}{\sqrt{2}}=\sqrt{2}\)
\(\Leftrightarrow A\ge\sqrt{2}+\sqrt{x}+2\ge\sqrt{2}+\sqrt{2}+2=2\sqrt{2}+2\)
Dấu "=" xảy ra khi: \(\dfrac{1}{2}\sqrt{x}=\dfrac{1}{\sqrt{x}}\Leftrightarrow x=2\)
Vậy, \(P\min\limits=2\sqrt{2}+2\Leftrightarrow x=2_{ }\)
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