Đáp án:
\(\begin{array}{l}
a)\dfrac{{x - \sqrt x + 1}}{{\sqrt x }}\\
b)\dfrac{3}{2}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)P = \dfrac{{\sqrt x - 1 + \sqrt x }}{{\sqrt x \left( {1 - \sqrt x } \right)}}:\left[ {\dfrac{{\left( {2\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\left( {1 - \sqrt x } \right)\left( {\sqrt x + 1} \right)}} + \dfrac{{\sqrt x \left( {2\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}} \right]\\
= \dfrac{{2\sqrt x - 1}}{{\sqrt x \left( {1 - \sqrt x } \right)}}:\left[ {\dfrac{{2\sqrt x - 1}}{{1 - \sqrt x }} + \dfrac{{\sqrt x \left( {2\sqrt x - 1} \right)}}{{x - \sqrt x + 1}}} \right]\\
= \dfrac{{2\sqrt x - 1}}{{\sqrt x \left( {1 - \sqrt x } \right)}}:\dfrac{{\left( {2\sqrt x - 1} \right)\left( {x - \sqrt x + 1} \right) + \left( {2x - \sqrt x } \right)\left( {1 - \sqrt x } \right)}}{{\left( {1 - \sqrt x } \right)\left( {x - \sqrt x + 1} \right)}}\\
= \dfrac{{2\sqrt x - 1}}{{\sqrt x \left( {1 - \sqrt x } \right)}}.\dfrac{{\left( {1 - \sqrt x } \right)\left( {x - \sqrt x + 1} \right)}}{{2x\sqrt x - 2x + 2\sqrt x - x + \sqrt x - 1 - 2x\sqrt x + 2x + x - \sqrt x }}\\
= \dfrac{{2\sqrt x - 1}}{{\sqrt x \left( {1 - \sqrt x } \right)}}.\dfrac{{\left( {1 - \sqrt x } \right)\left( {x - \sqrt x + 1} \right)}}{{2\sqrt x - 1}}\\
= \dfrac{{x - \sqrt x + 1}}{{\sqrt x }}\\
b)Thay:x = \dfrac{4}{{\sqrt {10} }}.\left( {\sqrt {3 + \sqrt 5 } + \sqrt {3 - \sqrt 5 } } \right)\\
= \dfrac{{4\left( {\sqrt {6 + 2\sqrt 5 } + \sqrt {6 - 2\sqrt 5 } } \right)}}{{2\sqrt 5 }}\\
= \dfrac{{2\left( {\sqrt {5 + 2\sqrt 5 .1 + 1} + \sqrt {5 - 2\sqrt 5 .1 + 1} } \right)}}{{\sqrt 5 }}\\
= \dfrac{{2\left( {\sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} } \right)}}{{\sqrt 5 }}\\
= \dfrac{{2\left( {\sqrt 5 + 1 + \sqrt 5 - 1} \right)}}{{\sqrt 5 }}\\
= \dfrac{{4\sqrt 5 }}{{\sqrt 5 }} = 4\\
\to P = \dfrac{{4 - \sqrt 4 + 1}}{{\sqrt 4 }} = \dfrac{{4 - 2 + 1}}{2} = \dfrac{3}{2}
\end{array}\)
