a/ ĐKXĐ: \(x>0;x\ne1\)
\(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right):\left(\dfrac{2}{x}-\dfrac{2-x}{x\sqrt{x}+x}\right)\)
= \(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{2\left(\sqrt{x}+1\right)-2+x}{x\left(\sqrt{x}+1\right)}\)
= \(\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{x+2\sqrt{x}}{x\left(\sqrt{x}+1\right)}\)
= \(\dfrac{x}{\sqrt{x}-1}\)
b/ Với \(x>0;x\ne1\)
Để P>2 \(\Leftrightarrow\dfrac{x}{\sqrt{x}-1}>2\Leftrightarrow\dfrac{x-2\sqrt{x}+2}{\sqrt{x}-1}>0\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}-1\right)^2+1}{\sqrt{x}-1}>0\)
Ta có: \(\left(\sqrt{x}-1\right)^2>0\) với mọi \(x>0,x\ne1\)
\(\Rightarrow\left(\sqrt{x}-1\right)^2+1>0\) với mọi x
Khi đó, \(\dfrac{\left(\sqrt{x}-1\right)^2+1}{\sqrt{x}-1}>0\) \(\Leftrightarrow\sqrt{x}-1>0\)
\(\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)
Vậy để P>2 thì x>1
c/ với \(x>0,x\ne1\)
Ta có: \(\dfrac{x}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)^2+1+2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
= \(\left(\sqrt{x}-1\right)+\dfrac{1}{\sqrt{x}-1}+2\)
Áp dụng bđt Co-si ta có:
\(\left(\sqrt{x}-1\right)+\dfrac{1}{\sqrt{x}-1}\ge2\sqrt{\left(\sqrt{x}-1\right).\dfrac{1}{\sqrt{x}-1}}\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)+\dfrac{1}{\sqrt{x}-1}\ge2\)
\(\Rightarrow\left(\sqrt{x}-1\right)+\dfrac{1}{\sqrt{x}-1}+2\ge4\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}-1=\dfrac{1}{\sqrt{x}-1}\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2=1\)
\(\Leftrightarrow x-2\sqrt{x}=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=4\left(tm\right)\end{matrix}\right.\)
Vậy GTNN của P là 4 khi x=4
