điều kiện xác định : \(a\ge0;a\ne1\)
a) ta có : \(P=\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)
\(\Leftrightarrow P=\left(\dfrac{a+\sqrt{a}+\sqrt{a}-1}{\sqrt{a}-1}\right)\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(\Leftrightarrow P=\left(\dfrac{a+2\sqrt{a}-1}{\sqrt{a}-1}\right)\left(1-\sqrt{a}\right)=1-2\sqrt{a}-a\)
b) ta có : \(P>-\sqrt{2}\Leftrightarrow1-2\sqrt{a}-a>-\sqrt{2}\)
\(\Leftrightarrow-\left(\sqrt{a}+1\right)^2\ge-\left(2+\sqrt{2}\right)\) \(\Leftrightarrow\left(\sqrt{a}+1\right)^2< \left(2+\sqrt{2}\right)\)
\(\Leftrightarrow-\sqrt{2+\sqrt{2}}< \sqrt{a}+1\le\sqrt{2+\sqrt{2}}\)
\(\Leftrightarrow-\sqrt{2+\sqrt{2}}-1\le\sqrt{a}\le\sqrt{2+\sqrt{2}}-1\)
\(\Leftrightarrow0\le\sqrt{a}< \sqrt{2+\sqrt{2}}-1\Leftrightarrow0\le a< 3+\sqrt{2}+2\sqrt{2+\sqrt{2}}\)
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c) ta có : \(P=\sqrt{a}\Leftrightarrow1-2\sqrt{a}-a=\sqrt{a}\Leftrightarrow-a-3\sqrt{a}+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{a}=\dfrac{-3+\sqrt{13}}{2}\left(N\right)\\\sqrt{a}=\dfrac{-3-\sqrt{13}}{2}\left(L\right)\end{matrix}\right.\)
ta có : \(\sqrt{a}=\dfrac{-3+\sqrt{13}}{2}\Rightarrow a=\dfrac{11-3\sqrt{13}}{2}\)
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