Đáp án:
c. \(MaxP = 1\)
Giải thích các bước giải:
\(\begin{array}{l}
C48:\\
a.DK:x>0;y \ge 0;x \ne y \ne 1\\
P = \left[ {\dfrac{{\left( {\sqrt x + \sqrt y } \right)\left( {1 + \sqrt {xy} } \right) + \left( {\sqrt x - \sqrt y } \right)\left( {1 - \sqrt {xy} } \right)}}{{\left( {1 - \sqrt {xy} } \right)\left( {1 + \sqrt {xy} } \right)}}} \right]:\left( {\dfrac{{1 - xy + x + y + 2xy}}{{\left( {1 - \sqrt {xy} } \right)\left( {1 + \sqrt {xy} } \right)}}} \right)\\
= \dfrac{{\sqrt x + \sqrt y + x\sqrt y + y\sqrt x + \sqrt x - \sqrt y - x\sqrt y + y\sqrt x }}{{\left( {1 - \sqrt {xy} } \right)\left( {1 + \sqrt {xy} } \right)}}.\dfrac{{\left( {1 - \sqrt {xy} } \right)\left( {1 + \sqrt {xy} } \right)}}{{x + y + xy + 1}}\\
= \dfrac{{2\sqrt x + 2y\sqrt x }}{{x + y + xy + 1}}\\
= \dfrac{{2\sqrt x \left( {y + 1} \right)}}{{x\left( {y + 1} \right) + y + 1}}\\
= \dfrac{{2\sqrt x }}{{x + 1}}\\
b.Thay:x = \dfrac{2}{{2 + \sqrt 3 }} = \dfrac{4}{{4 + 2\sqrt 3 }} = \dfrac{4}{{{{\left( {\sqrt 3 + 1} \right)}^2}}}\\
\to P = \dfrac{{2\sqrt {\dfrac{4}{{{{\left( {\sqrt 3 + 1} \right)}^2}}}} }}{{\dfrac{2}{{2 + \sqrt 3 }} + 1}} = \dfrac{{2.\dfrac{2}{{\sqrt 3 + 1}}}}{{\dfrac{{2 + 2 + \sqrt 3 }}{{2 + \sqrt 3 }}}}\\
= \dfrac{4}{{\sqrt 3 + 1}}.\dfrac{{2 + \sqrt 3 }}{{4 + \sqrt 3 }} = \dfrac{{2 + 6\sqrt 3 }}{{13}}
\end{array}\)
c. Để P đạt GTLN
⇔ \(\dfrac{1}{P}\) đạt GTNN
Có:
\(\begin{array}{l}
\dfrac{1}{P} = \dfrac{{x + 1}}{{2\sqrt x }} = \dfrac{{\sqrt x }}{2} + \dfrac{1}{{2\sqrt x }}\\
Do:x > 0\\
Co - si \to \dfrac{{\sqrt x }}{2} + \dfrac{1}{{2\sqrt x }} \ge 2\sqrt {\dfrac{{\sqrt x }}{2}.\dfrac{1}{{2\sqrt x }}} = 2.\dfrac{1}{2} = 1\\
\to Min\dfrac{1}{P} = 1\\
\to MaxP = 1\\
\Leftrightarrow x = 1
\end{array}\)
