Đáp án:
b) \(P = - \sqrt 2 - 1\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0;x \ne \dfrac{1}{2}\\
P = \left[ {\dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt {2x} - 1} \right) + \left( {\sqrt {2x} + \sqrt x } \right)\left( {\sqrt {2x} + 1} \right) - 2x + 1}}{{\left( {\sqrt {2x} - 1} \right)\left( {\sqrt {2x} + 1} \right)}}} \right]:\dfrac{{2x - 1 + \left( {\sqrt x + 1} \right)\left( {\sqrt {2x} - 1} \right) - \left( {\sqrt {2x} + \sqrt x } \right)\left( {\sqrt {2x} + 1} \right)}}{{\left( {\sqrt {2x} - 1} \right)\left( {\sqrt {2x} + 1} \right)}}\\
= \dfrac{{x\sqrt 2 - \sqrt x + \sqrt {2x} - 1 + 2x + \sqrt {2x} + x\sqrt 2 + \sqrt x - 2x + 1}}{{\left( {\sqrt {2x} - 1} \right)\left( {\sqrt {2x} + 1} \right)}}.\dfrac{{\left( {\sqrt {2x} - 1} \right)\left( {\sqrt {2x} + 1} \right)}}{{2x - 1 + x\sqrt 2 - \sqrt x + \sqrt {2x} - 1 - 2x - \sqrt {2x} - x\sqrt 2 - \sqrt x }}\\
= \dfrac{{2x\sqrt 2 + 2\sqrt {2x} }}{{ - 2 - 2\sqrt x }}\\
= \dfrac{{x\sqrt 2 + \sqrt {2x} }}{{ - 1 - \sqrt x }} = \dfrac{{\sqrt {2x} \left( {\sqrt x + 1} \right)}}{{ - \left( {\sqrt x + 1} \right)}} = - \sqrt {2x} \\
b)Thay:x = \dfrac{1}{2}\left( {3 + 2\sqrt 2 } \right)\\
= \dfrac{1}{2}.\left( {2 + 2\sqrt 2 .1 + 1} \right)\\
= \dfrac{1}{2}{\left( {\sqrt 2 + 1} \right)^2}\\
\to P = - \sqrt {2.\dfrac{1}{2}{{\left( {\sqrt 2 + 1} \right)}^2}} = - \sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} \\
= - \sqrt 2 - 1
\end{array}\)
