a)\(P=\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}+\dfrac{\sqrt{a}-1}{\sqrt{a}+1}\right)\) đk\(\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\)
\(=\dfrac{a+\sqrt{a}+1}{\sqrt{a}}-\dfrac{a-\sqrt{a}+1}{\sqrt{a}}+\left(\dfrac{a-1}{\sqrt{a}}\right)\left(\dfrac{a+2\sqrt{a}+1+a-2\sqrt{a}+1}{a-1}\right)\)
\(=\dfrac{2\sqrt{a}}{\sqrt{a}}+\left(\dfrac{a-1}{\sqrt{a}}\right)\left(\dfrac{2a+2}{a-1}\right)\)
\(=2+\dfrac{2a+2}{\sqrt{a}}=\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}\)
b) Ta có P=7
=>\(\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}=7\)
\(\Leftrightarrow2a+2\sqrt{a}+2=7\sqrt{a}\)
\(\Leftrightarrow2a-5\sqrt{a}+2=0\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{a}-2=0\\2\sqrt{a}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=4\left(n\right)\\a=\dfrac{1}{4}\left(n\right)\end{matrix}\right.\)
Vậy \(a=\left\{4,\dfrac{1}{4}\right\}\) Thì P=7
c) Ta có P =\(\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}=2\sqrt{a}+2+\dfrac{2}{\sqrt{a}}\)
Áp dụng BĐT Cauchy cho 2 số \(2\sqrt{a}+\dfrac{2}{\sqrt{a}}\ge2\sqrt{2\sqrt{a}.\dfrac{2}{\sqrt{a}}}=2\sqrt{4}=4\)
=> \(P\ge4+2=6\)
Dấu "=" xảy ra \(\Leftrightarrow2\sqrt{a}=\dfrac{2}{\sqrt{a}}\Rightarrow a=1\left(loạivìa\ne1\right)\)=> Dấu "=" không xảy ra
=>\(P>6\left(đpcm\right)\)
