Question

Đã trả lời bởi chuyên gia

Answer 1

\[\begin{array}{l}
P = \left( {\frac{{\sqrt x + \sqrt y }}{{1 - \sqrt {xy} }} + \frac{{\sqrt x - \sqrt y }}{{1 + \sqrt {xy} }}} \right) \div \left( {1 + \frac{{x + y + 2xy}}{{1 - xy}}} \right)\\
dk:x;y \ge 0;xy \ne 1\\
= \left[ {\frac{{\left( {\sqrt x + \sqrt y } \right)\left( {1 + \sqrt {xy} } \right) + \left( {\sqrt x - \sqrt y } \right)\left( {1 - \sqrt {xy} } \right)}}{{\left( {1 - \sqrt {xy} } \right)\left( {1 + \sqrt {xy} } \right)}}} \right]:\left[ {\frac{{1 - xy + x + y + 2xy}}{{\left( {1 - \sqrt {xy} } \right)\left( {1 + \sqrt {xy} } \right)}}} \right]\\
= \frac{{\sqrt x + x\sqrt y + \sqrt y + y\sqrt x + \sqrt x - x\sqrt y - \sqrt y + y\sqrt x }}{{\left( {1 - \sqrt {xy} } \right)\left( {1 + \sqrt {xy} } \right)}}:\frac{{xy + x + y + 1}}{{\left( {1 - \sqrt {xy} } \right)\left( {1 + \sqrt {xy} } \right)}}\\
= \frac{{2\sqrt x + 2y\sqrt x }}{{\left( {1 - \sqrt {xy} } \right)\left( {1 + \sqrt {xy} } \right)}}.\frac{{\left( {1 - \sqrt {xy} } \right)\left( {1 + \sqrt {xy} } \right)}}{{x\left( {y + 1} \right) + \left( {y + 1} \right)}}\\
= \frac{{2\sqrt x \left( {1 + y} \right)}}{{\left( {y + 1} \right)\left( {x + 1} \right)}}\\
= \frac{{2\sqrt x }}{{x + 1}}
\end{array}\]

có: \[x = \frac{2}{{2 + \sqrt 3 }} = \frac{{2\left( {2 - \sqrt 3 } \right)}}{{{2^2} - 3}} = 4 - 2\sqrt 3 = 3 - 2\sqrt 3 + 1 = {\left( {\sqrt 3 - 1} \right)^2}\]

\[ = > \sqrt x = \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} = |\sqrt 3 - 1| = \sqrt 3 - 1\]

thay vào P ta có:

\[\begin{array}{l}
P = \frac{{2.\left( {\sqrt 3 - 1} \right)}}{{\frac{2}{{2 + \sqrt 3 }} + 1}} = \frac{{\left( {2\sqrt 3 - 2} \right)\left( {2 + \sqrt 3 } \right)}}{{2 + 2 + \sqrt 3 }}\\
= \frac{{4\sqrt 3 + 6 - 4 - 2\sqrt 3 }}{{4 + \sqrt 3 }} = \frac{{2 + 2\sqrt 3 }}{{4 + \sqrt 3 }}\\
= \frac{{\left( {2 + 2\sqrt 3 } \right)\left( {4 - \sqrt 3 } \right)}}{{16 - 3}}\\
= \frac{{8 - 2\sqrt 3 + 8\sqrt 3 - 6}}{{13}}\\
= \frac{{2 + 6\sqrt 3 }}{{13}}
\end{array}\]

Vậy với \[x = \frac{2}{{2 + \sqrt 3 }}\] thì \[P = \frac{{2 + 6\sqrt 3 }}{{13}}\]

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Answer 2

\[\begin{array}{l}
P = \left( {\frac{{\sqrt x + \sqrt y }}{{1 - \sqrt {xy} }} + \frac{{\sqrt x - \sqrt y }}{{1 + \sqrt {xy} }}} \right) \div \left( {1 + \frac{{x + y + 2xy}}{{1 - xy}}} \right)\\
dk:x;y \ge 0;xy \ne 1\\
= \left[ {\frac{{\left( {\sqrt x + \sqrt y } \right)\left( {1 + \sqrt {xy} } \right) + \left( {\sqrt x - \sqrt y } \right)\left( {1 - \sqrt {xy} } \right)}}{{\left( {1 - \sqrt {xy} } \right)\left( {1 + \sqrt {xy} } \right)}}} \right]:\left[ {\frac{{1 - xy + x + y + 2xy}}{{\left( {1 - \sqrt {xy} } \right)\left( {1 + \sqrt {xy} } \right)}}} \right]\\
= \frac{{\sqrt x + x\sqrt y + \sqrt y + y\sqrt x + \sqrt x - x\sqrt y - \sqrt y + y\sqrt x }}{{\left( {1 - \sqrt {xy} } \right)\left( {1 + \sqrt {xy} } \right)}}:\frac{{xy + x + y + 1}}{{\left( {1 - \sqrt {xy} } \right)\left( {1 + \sqrt {xy} } \right)}}\\
= \frac{{2\sqrt x + 2y\sqrt x }}{{\left( {1 - \sqrt {xy} } \right)\left( {1 + \sqrt {xy} } \right)}}.\frac{{\left( {1 - \sqrt {xy} } \right)\left( {1 + \sqrt {xy} } \right)}}{{x\left( {y + 1} \right) + \left( {y + 1} \right)}}\\
= \frac{{2\sqrt x \left( {1 + y} \right)}}{{\left( {y + 1} \right)\left( {x + 1} \right)}}\\
= \frac{{2\sqrt x }}{{x + 1}}
\end{array}\]

có: \[x = \frac{2}{{2 + \sqrt 3 }} = \frac{{2\left( {2 - \sqrt 3 } \right)}}{{{2^2} - 3}} = 4 - 2\sqrt 3 = 3 - 2\sqrt 3 + 1 = {\left( {\sqrt 3 - 1} \right)^2}\]

\[ = > \sqrt x = \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} = |\sqrt 3 - 1| = \sqrt 3 - 1\]

thay vào P ta có:

\[\begin{array}{l}
P = \frac{{2.\left( {\sqrt 3 - 1} \right)}}{{\frac{2}{{2 + \sqrt 3 }} + 1}} = \frac{{\left( {2\sqrt 3 - 2} \right)\left( {2 + \sqrt 3 } \right)}}{{2 + 2 + \sqrt 3 }}\\
= \frac{{4\sqrt 3 + 6 - 4 - 2\sqrt 3 }}{{4 + \sqrt 3 }} = \frac{{2 + 2\sqrt 3 }}{{4 + \sqrt 3 }}\\
= \frac{{\left( {2 + 2\sqrt 3 } \right)\left( {4 - \sqrt 3 } \right)}}{{16 - 3}}\\
= \frac{{8 - 2\sqrt 3 + 8\sqrt 3 - 6}}{{13}}\\
= \frac{{2 + 6\sqrt 3 }}{{13}}
\end{array}\]

Vậy với \[x = \frac{2}{{2 + \sqrt 3 }}\] thì \[P = \frac{{2 + 6\sqrt 3 }}{{13}}\]

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