Xét P = \(\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}\) \(\left(x>0,x\ne1\right)\)
\(\Leftrightarrow\) P = \(\dfrac{x-\sqrt{x}+2\sqrt{x}-2+3}{\sqrt{x}-1}\)
\(\Leftrightarrow\) P = \(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}+\dfrac{2\sqrt{x}-2}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}-1}\)
\(\Leftrightarrow\) P = \(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}-1}\)
\(\Leftrightarrow\) P = \(\sqrt{x}+2+\dfrac{3}{\sqrt{x}-1}\)
\(\Leftrightarrow\) P = \(\sqrt{x}-1+\dfrac{3}{\sqrt{x}-1}+3\)
Ta có : \(x>0,x\ne1\Leftrightarrow x>1\Leftrightarrow\sqrt{x}>1\Leftrightarrow\sqrt{x}-1>0\Leftrightarrow\dfrac{3}{\sqrt{x}-1}>0\)
Áp dụng bất đẳng thức Cô - si cho hai số dương \(\sqrt{x}-1\) và \(\dfrac{3}{\sqrt{x}-1}\) ta được:
\(\sqrt{x}-1+\dfrac{3}{\sqrt{x}-1}\ge2\sqrt{\left(\sqrt{x}-1\right).\left(\dfrac{3}{\sqrt{x}-1}\right)}\)
\(\Leftrightarrow\) \(\sqrt{x}-1+\dfrac{3}{\sqrt{x}-1}+3\ge2\sqrt{\left(\sqrt{x}-1\right).\left(\dfrac{3}{\sqrt{x}-1}\right)}+3\)
\(\Leftrightarrow\) P \(\ge2\sqrt{3}+3\)
Dấu " = " xảy ra \(\Leftrightarrow\sqrt{x}-1=\dfrac{3}{\sqrt{x}-1}\Leftrightarrow\left(\sqrt{x}-1\right)^2=3\Leftrightarrow\sqrt{x}-1=\pm\sqrt{3}\)
*TH1: \(\sqrt{x}-1=\sqrt{3}\Leftrightarrow\sqrt{x}=1+\sqrt{3}\) \(\Leftrightarrow x=\left(1+\sqrt{3}\right)^2\Leftrightarrow x=4+2\sqrt{3}\) (thỏa mãn điều kiện )
*TH2: \(\sqrt{x}-1=-\sqrt{3}\Leftrightarrow\sqrt{x}=1-\sqrt{3}\) (vô lý)
Vậy giá trị nhỏ nhất của P = \(\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}\) là \(2\sqrt{3}+3\) khi \(x=4+2\sqrt{3}\)
ĐKXĐ: x ≥ 0; x ≠ 1
+) Xét 0 ≤ x < 1 => \(\left\{{}\begin{matrix}x+\sqrt{x}+1>0\\\sqrt{x}-1< 0\end{matrix}\right.\) => P < 0
+) Xét x > 1
Ta có: \(P-1=\dfrac{x+2}{\sqrt{x}-1}=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{2\sqrt{x}-2}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+2+\dfrac{3}{\sqrt{x}-1}=\left(\sqrt{x}-1+\dfrac{3}{\sqrt{x}-1}\right)+2\ge2\sqrt{3}+2\)
=> \(P\ge2\sqrt{3}+3\)
Dấu "=" xảy ra <=> \(\sqrt{x}-1=\dfrac{3}{\sqrt{x}-1}\Leftrightarrow x=\left(\sqrt{3}+1\right)^2\)
