Question

giúp mình gấp với

Answer 1

a: \(P=\dfrac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}+1\right)}\cdot\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}-1}=\dfrac{\sqrt{a}+1}{\sqrt{a}}\)

Answer 2

a) \(P=\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{1}{a+\sqrt{a}}\right):\dfrac{\sqrt{a}-1}{a+2\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}+1\right)}.\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}-1}=\dfrac{\sqrt{a}+1}{\sqrt{a}}\)

b) P > 2

\(\Leftrightarrow\dfrac{\sqrt{a}+1}{\sqrt{a}}>2\Leftrightarrow\sqrt{a}+1>2\sqrt{a}\Leftrightarrow\sqrt{a}< 1\Leftrightarrow0< a< 1\)