Ta có:
\(P=\dfrac{x}{\sqrt{x}-1}=\dfrac{x-4\sqrt{x}+4+4\sqrt{x}-4}{\sqrt{x}-1}\)
\(P=\dfrac{\left(x-4\sqrt{x}+4\right)+4\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-1}+4\)
Mà: \(x>1\Rightarrow\sqrt{x}-1>0\Rightarrow\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-1}\ge0\forall x\)
\(\Rightarrow P=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-1}+4\ge4\forall x\)
Dấu "=" xảy ra:
\(\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-1}+4=4\)
\(\Leftrightarrow\sqrt{x}-2=0\)
\(\Leftrightarrow\sqrt{x}=2\)
\(\Leftrightarrow x=4\)
Vậy: \(P_{min}=4\Leftrightarrow x=4\)
\(P=\dfrac{x}{\sqrt[]{x}-1}\left(x>1\right)\)
\(\Leftrightarrow P=\dfrac{x-1+1}{\sqrt[]{x}-1}\)
\(\Leftrightarrow P=\dfrac{\left(\sqrt[]{x}-1\right)\left(\sqrt[]{x}+1\right)+1}{\sqrt[]{x}-1}\)
\(\Leftrightarrow P=\sqrt[]{x}+1+\dfrac{1}{\sqrt[]{x}-1}\)
\(\Leftrightarrow P=\sqrt[]{x}-1+\dfrac{1}{\sqrt[]{x}-1}+2\)
Áp dụng BĐT Cauchy cho 2 số dương \(\sqrt[]{x}-1;\dfrac{1}{\sqrt[]{x}-1}\)
\(P=\sqrt[]{x}-1+\dfrac{1}{\sqrt[]{x}-1}+2\ge2+2=4\)
Dấu "=" xảy ra khi và chỉ khi
\(\sqrt[]{x}-1=\dfrac{1}{\sqrt[]{x}-1}\)
\(\Leftrightarrow\left(\sqrt[]{x}-1\right)^2=1\)
\(\Leftrightarrow\sqrt[]{x}-1=1\left(x>1\Rightarrow\sqrt[]{x}-1>0\right)\)
\(\Leftrightarrow\sqrt[]{x}=2\)
\(\Leftrightarrow x=4\)
Vậy \(GTNN\left(P\right)=4\left(khi.x=4\right)\)
