a, ĐK: \(a\ge0;a\ne1\)
\(P=\left(\dfrac{1}{a+\sqrt{a}}+\dfrac{1}{\sqrt{a}+1}\right):\dfrac{\sqrt{a}-1}{a+2\sqrt{a}+1}\)
\(=\left[\dfrac{1}{\sqrt{a}\left(\sqrt{a}+1\right)}+\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}+1\right)}\right]:\dfrac{\sqrt{a}-1}{\left(\sqrt{a}+1\right)^2}\)
\(=\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}.\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}-1}\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2}{a-\sqrt{a}}\)
b, \(P=\dfrac{\left(\sqrt{a}+1\right)^2}{a-\sqrt{a}}=\dfrac{1}{2}\)
\(\Leftrightarrow2\left(\sqrt{a}+1\right)^2=a-\sqrt{a}\)
\(\Leftrightarrow2\left(a+2\sqrt{a}+1\right)=a-\sqrt{a}\)
\(\Leftrightarrow a+5\sqrt{a}+2=0\)
\(\Rightarrow\) vô nghiệm.
Vậy không tồn tại giá trị a thỏa mãn \(P=\dfrac{1}{2}\).
