Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}
B5:\\
a.DK:x > 0;x \ne 1\\
P = \left( {\frac{{\sqrt x - 1 + \sqrt x }}{{\sqrt x \left( {1 - \sqrt x } \right)}}} \right):\left( {\frac{{\left( {2\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}} + \frac{{\sqrt x \left( {2\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}} \right)\\
= \frac{{2\sqrt x - 1}}{{\sqrt x \left( {1 - \sqrt x } \right)}}:\left[ {\frac{{2\sqrt x - 1}}{{1 - \sqrt x }} + \frac{{\sqrt x \left( {2\sqrt x - 1} \right)}}{{x - \sqrt x + 1}}} \right]\\
= \frac{{2\sqrt x - 1}}{{\sqrt x \left( {1 - \sqrt x } \right)}}:\frac{{2x\sqrt x - 2x + 2\sqrt x - x + \sqrt x - 1 + 2x - \sqrt x - 2x\sqrt x + x}}{{\left( {1 - \sqrt x } \right)\left( {x - \sqrt x + 1} \right)}}\\
= \frac{{2\sqrt x - 1}}{{\sqrt x \left( {1 - \sqrt x } \right)}}.\frac{{\left( {1 - \sqrt x } \right)\left( {x - \sqrt x + 1} \right)}}{{2\sqrt x - 1}}\\
= \frac{{x - \sqrt x + 1}}{{\sqrt x }}\\
b.x = 7 - 4\sqrt 3 \\
\to P = \frac{{7 - 4\sqrt 3 - \sqrt {7 - 4\sqrt 3 } + 1}}{{\sqrt {7 - 4\sqrt 3 } }} = 3\\
c.P = \frac{{x + 1 - \sqrt x }}{{\sqrt x }} = \sqrt x + \frac{1}{{\sqrt x }} - 1 \ge 2\sqrt {\sqrt x .\frac{1}{{\sqrt x }}} - 1 = 2 - 1 = 1\left( {Co - si} \right)\\
'' = '' \Leftrightarrow \sqrt x = \frac{1}{{\sqrt x }} \Leftrightarrow x = 1
\end{array}\)
