Đáp án:
c. \(\dfrac{{ - 1 + \sqrt 2 }}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:a > );a \ne \left\{ {1;4} \right\}\\
b.P = \left[ {\dfrac{{\sqrt a - \sqrt a + 1}}{{\sqrt a \left( {\sqrt a - 1} \right)}}} \right]:\left[ {\dfrac{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right) - \left( {\sqrt a + 2} \right)\left( {\sqrt a - 2} \right)}}{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 2} \right)}}} \right]\\
= \dfrac{1}{{\sqrt a \left( {\sqrt a - 1} \right)}}.\dfrac{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 2} \right)}}{{a - 1 - a + 4}}\\
= \dfrac{1}{{\sqrt a \left( {\sqrt a - 1} \right)}}.\dfrac{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 2} \right)}}{3}\\
= \dfrac{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 2} \right)}}{{3\sqrt a \left( {\sqrt a - 1} \right)}}\\
c.P = \dfrac{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 2} \right)}}{{3\sqrt a \left( {\sqrt a - 1} \right)}}\\
= \dfrac{{a - \sqrt a - 2}}{{3a - 3\sqrt a }}\\
Thay:a = 3 + 2\sqrt 2 = {\left( {\sqrt 2 } \right)^2} + 2\sqrt 2 + 1\\
= {\left( {\sqrt 2 + 1} \right)^2}\\
\to \sqrt a = \sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} = \sqrt 2 + 1\\
\to P = \dfrac{{3 + 2\sqrt 2 - \sqrt 2 - 1 - 2}}{{9 + 6\sqrt 2 - 3\sqrt 2 - 3}} = \\
= \dfrac{{\sqrt 2 }}{{6 + 3\sqrt 2 }} = \dfrac{{ - 1 + \sqrt 2 }}{3}
\end{array}\)
