Đáp án:
a) \(\sqrt {\dfrac{x}{y}} \)
b) \(P = \sqrt 2 \)
Giải thích các bước giải:
\(\begin{array}{l}
a)P = \dfrac{x}{{\sqrt y \left( {\sqrt x - 2\sqrt y } \right)}} - \dfrac{{2\sqrt x }}{{\sqrt x \left( {\sqrt x + 1} \right) - 2\sqrt y \left( {\sqrt x + 1} \right)}}.\dfrac{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}{{1 - \sqrt x }}\\
= \dfrac{x}{{\sqrt y \left( {\sqrt x - 2\sqrt y } \right)}} - \dfrac{{2\sqrt x }}{{\left( {\sqrt x - 2\sqrt y } \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}{{1 - \sqrt x }}\\
= \dfrac{x}{{\sqrt y \left( {\sqrt x - 2\sqrt y } \right)}} - \dfrac{{2\sqrt x }}{{\sqrt x - 2\sqrt y }}\\
= \dfrac{{x - 2\sqrt {xy} }}{{\sqrt y \left( {\sqrt x - 2\sqrt y } \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 2\sqrt y } \right)}}{{\sqrt y \left( {\sqrt x - 2\sqrt y } \right)}}\\
= \sqrt {\dfrac{x}{y}} \\
b)2{x^2} + 4{y^2} - 6x - 4xy + 9 = 0\\
\to {x^2} - 2.x.2y + 4{y^2} + {x^2} - 6x + 9 = 0\\
\to {\left( {x - 2y} \right)^2} + {\left( {x - 3} \right)^2} = 0\\
\to \left\{ \begin{array}{l}
x - 3 = 0\\
x - 2y = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 3\\
y = \dfrac{3}{2}
\end{array} \right.\\
Thay:\left\{ \begin{array}{l}
x = 3\\
y = \dfrac{3}{2}
\end{array} \right.\\
\to P = \sqrt {\dfrac{3}{{\dfrac{3}{2}}}} = \sqrt 2
\end{array}\)
