Đáp án:
a) \(\dfrac{{\sqrt {xy} }}{{x - \sqrt {xy} + y}}\)
Giải thích các bước giải:
Bài 7:
\(\begin{array}{l}
a)DK:x > 0;y > 0;x \ne y\\
P = \left[ {\dfrac{{\left( {\sqrt x - \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right)}}{{\sqrt x - \sqrt y }} + \dfrac{{\left( {\sqrt x - \sqrt y } \right)\left( {x + \sqrt {xy} + y} \right)}}{{ - \left( {\sqrt x - \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right)}}} \right].\dfrac{{\sqrt x + \sqrt y }}{{x - 2\sqrt {xy} + y + \sqrt {xy} }}\\
= \left( {\sqrt x + \sqrt y - \dfrac{{x + \sqrt {xy} + y}}{{\sqrt x + \sqrt y }}} \right).\dfrac{{\sqrt x + \sqrt y }}{{x - \sqrt {xy} + y}}\\
= \dfrac{{x + 2\sqrt {xy} + y - x - \sqrt {xy} - y}}{{\sqrt x + \sqrt y }}.\dfrac{{\sqrt x + \sqrt y }}{{x - \sqrt {xy} + y}}\\
= \dfrac{{\sqrt {xy} }}{{x - \sqrt {xy} + y}}\\
b)x - \sqrt {xy} + y = x - 2\sqrt x .\dfrac{1}{2}\sqrt y + \dfrac{1}{4}y + \dfrac{3}{4}y\\
= {\left( {\sqrt x - \dfrac{1}{2}\sqrt y } \right)^2} + \dfrac{3}{4}y > 0\forall x \ge 0;y \ge 0;x \ne y\\
Do:\left\{ \begin{array}{l}
\sqrt {xy} \ge 0\forall x \ge 0;y \ge 0;x \ne y\\
x - \sqrt {xy} + y > 0\forall x \ge 0;y \ge 0;x \ne y
\end{array} \right.\\
\to P \ge 0\forall x \ge 0;y \ge 0;x \ne y
\end{array}\)
