ĐKXĐ: a>0 và a<>1
a: \(P=\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}+\dfrac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)
\(=\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}+\dfrac{a-1}{\sqrt{a}}\cdot\dfrac{\left(\sqrt{a}+1\right)^2+\left(\sqrt{a}-1\right)^2}{a-1}\)
\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}+\dfrac{2a+2}{\sqrt{a}}\)
\(=\dfrac{2a+2+2\sqrt{a}}{\sqrt{a}}\)
b: P=7
=>\(2a+2\sqrt{a}+2=7\sqrt{a}\)
=>\(2a-5\sqrt{a}+2=0\)
=>\(\left(2\sqrt{a}-1\right)\left(\sqrt{a}-2\right)=0\)
=>a=1/4 hoặc a=4
c: \(P-6=\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}-6\)
\(=\dfrac{2a-4\sqrt{a}+2}{\sqrt{a}}=\dfrac{2\left(a-2\sqrt{a}+1\right)}{\sqrt{a}}\)
\(=\dfrac{2\left(\sqrt{a}-1\right)^2}{\sqrt{a}}>0\)
=>P>6
