Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
B = \left( {\dfrac{{16\sqrt y }}{{x - \sqrt {xy} }} - \dfrac{{17\sqrt x }}{{\sqrt {xy} - y}}} \right):\left( {\dfrac{1}{{\sqrt x }} + \dfrac{1}{{\sqrt y }}} \right)\\
= \left( {\dfrac{{16\sqrt y }}{{\sqrt x .\left( {\sqrt x - \sqrt y } \right)}} - \dfrac{{17\sqrt x }}{{\sqrt y \left( {\sqrt x - \sqrt y } \right)}}} \right):\dfrac{{\sqrt x + \sqrt y }}{{\sqrt {xy} }}\\
= \dfrac{{16\sqrt y .\sqrt y - 17\sqrt x .\sqrt x }}{{\sqrt x .\sqrt y .\left( {\sqrt x - \sqrt y } \right)}}:\dfrac{{\sqrt x + \sqrt y }}{{\sqrt {xy} }}\\
= \dfrac{{16y - 17x}}{{\sqrt {xy} \left( {\sqrt x - \sqrt y } \right)}}.\dfrac{{\sqrt {xy} }}{{\left( {\sqrt x + \sqrt y } \right)}}\\
= \dfrac{{16y - 17x}}{{\left( {\sqrt x - \sqrt y } \right).\left( {\sqrt x + \sqrt y } \right)}}\\
= \dfrac{{16y - 17x}}{{x - y}}\\
b,\\
x\left( {x + 2y} \right) = 8{y^2}\\
\Leftrightarrow {x^2} + 2xy - 8{y^2} = 0\\
\Leftrightarrow \left( {{x^2} + 4xy} \right) - \left( {2xy + 8{y^2}} \right) = 0\\
\Leftrightarrow x.\left( {x + 4y} \right) - 2y\left( {x + 4y} \right) = 0\\
\Leftrightarrow \left( {x + 4y} \right)\left( {x - 2y} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 4y\\
x = 2y
\end{array} \right.\\
x,y > 0 \Rightarrow x = 2y\\
\Rightarrow B = \dfrac{{16y - 17x}}{{x - y}} = \dfrac{{16y - 17.2y}}{{2y - y}} = \dfrac{{ - 18y}}{y} = - 18
\end{array}\)
