Đáp án:
a) \(A = \dfrac{{21}}{8}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Do:\dfrac{x}{y} = \dfrac{2}{3} \to x = \dfrac{2}{3}y\\
Thay:x = \dfrac{2}{3}y\\
\to A = \dfrac{{3.\dfrac{2}{3}y + 5y}}{{7.\dfrac{2}{3}y - 2y}} = \dfrac{{7y}}{{\dfrac{8}{3}y}} = \dfrac{{21}}{8}\\
b)Thay:x = \dfrac{2}{3}y\\
B = \dfrac{{{{\left( {\dfrac{2}{3}y} \right)}^2} - \dfrac{2}{3}y.y + {y^2}}}{{{{\left( {\dfrac{2}{3}y} \right)}^2} + \dfrac{2}{3}y.y + {y^2}}}\\
= \left( {\dfrac{4}{9}{y^2} - \dfrac{2}{3}{y^2} + {y^2}} \right):\left( {\dfrac{4}{9}{y^2} + \dfrac{2}{3}{y^2} + {y^2}} \right)\\
= \dfrac{7}{9}{y^2}:\dfrac{{19}}{9}{y^2}\\
= \dfrac{7}{{19}}
\end{array}\)
