Đáp án:
\(\begin{array}{l}
a,\\
P = \dfrac{{\sqrt a }}{{\sqrt a + 1}}\\
b,\\
P = \dfrac{1}{3}
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
P = \left( {1 + \dfrac{1}{{\sqrt a - 1}}} \right):\left( {\dfrac{{a + \sqrt a }}{{a - 1}} + \dfrac{1}{{\sqrt a - 1}}} \right)\\
= \dfrac{{\left( {\sqrt a - 1} \right) + 1}}{{\sqrt a - 1}}:\left( {\dfrac{{{{\sqrt a }^2} + \sqrt a }}{{{{\sqrt a }^2} - {1^2}}} + \dfrac{1}{{\sqrt a - 1}}} \right)\\
= \dfrac{{\sqrt a }}{{\sqrt a - 1}}:\left( {\dfrac{{\sqrt a \left( {\sqrt a + 1} \right)}}{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}} + \dfrac{1}{{\sqrt a - 1}}} \right)\\
= \dfrac{{\sqrt a }}{{\sqrt a - 1}}:\left( {\dfrac{{\sqrt a }}{{\sqrt a - 1}} + \dfrac{1}{{\sqrt a - 1}}} \right)\\
= \dfrac{{\sqrt a }}{{\sqrt a - 1}}:\dfrac{{\sqrt a + 1}}{{\sqrt a - 1}}\\
= \dfrac{{\sqrt a }}{{\sqrt a - 1}}.\dfrac{{\sqrt a - 1}}{{\sqrt a + 1}}\\
= \dfrac{{\sqrt a }}{{\sqrt a + 1}}\\
b,\\
a = \dfrac{1}{4} \Rightarrow \sqrt a = \dfrac{1}{2} \Rightarrow P = \dfrac{{\dfrac{1}{2}}}{{\dfrac{1}{2} + 1}} = \dfrac{{\dfrac{1}{2}}}{{\dfrac{3}{2}}} = \dfrac{1}{3}
\end{array}\)