Sửa đề: \(P=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{a-1}+\dfrac{\sqrt{a}}{\sqrt{a}+1}\right):\dfrac{\sqrt{a}}{\sqrt{a}+1}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
a) Ta có: \(P=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{a-1}+\dfrac{\sqrt{a}}{\sqrt{a}+1}\right):\dfrac{\sqrt{a}}{\sqrt{a}+1}\)
\(=\left(\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\dfrac{2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right):\dfrac{\sqrt{a}}{\sqrt{a}+1}\)
\(=\dfrac{\sqrt{a}+1-2\sqrt{a}+a-\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}:\dfrac{\sqrt{a}}{\sqrt{a}+1}\)
\(=\dfrac{a-2\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}+1}{\sqrt{a}}\)
\(=\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}\left(\sqrt{a}-1\right)}\)
\(=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
b) Để \(P< \dfrac{1}{2}\) thì \(P-\dfrac{1}{2}< 0\)
\(\Leftrightarrow\dfrac{\sqrt{a}-1}{\sqrt{a}}-\dfrac{1}{2}< 0\)
\(\Leftrightarrow\dfrac{2\left(\sqrt{a}-1\right)}{2\sqrt{a}}-\dfrac{\sqrt{a}}{2\sqrt{a}}< 0\)
\(\Leftrightarrow\dfrac{2\sqrt{a}-2-\sqrt{a}}{2\sqrt{a}}< 0\)
\(\Leftrightarrow\dfrac{\sqrt{a}-2}{2\sqrt{a}}< 0\)
mà \(2\sqrt{a}>0\forall a\) thỏa mãn ĐKXĐ
nên \(\sqrt{a}-2< 0\)
\(\Leftrightarrow\sqrt{a}< 2\)
hay a<4
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0< a< 4\\a\ne1\end{matrix}\right.\)
Vậy: Để \(P< \dfrac{1}{2}\) thì \(\left\{{}\begin{matrix}0< a< 4\\a\ne1\end{matrix}\right.\)
