Giải thích các bước giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\begin{array}{l}
a,\\
\dfrac{x}{4} = \dfrac{y}{7} \Leftrightarrow \dfrac{{2.x}}{{2.4}} = \dfrac{{3.y}}{{3.7}} \Leftrightarrow \dfrac{{2x}}{8} = \dfrac{{3y}}{{21}} = \dfrac{{2x - 3y}}{{8 - 21}} = \dfrac{{39}}{{ - 13}} = - 3\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{{2x}}{8} = - 3\\
\dfrac{{3y}}{{21}} = - 3
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{x}{4} = - 3\\
\dfrac{y}{7} = - 3
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = - 12\\
y = - 21
\end{array} \right.\\
b,\\
\dfrac{a}{b} = \dfrac{c}{d} \Leftrightarrow {\left( {\dfrac{a}{b}} \right)^2} = {\left( {\dfrac{c}{d}} \right)^2} \Leftrightarrow \dfrac{{{a^2}}}{{{b^2}}} = \dfrac{{{c^2}}}{{{d^2}}} = \dfrac{{{a^2} + {c^2}}}{{{b^2} + {d^2}}}\\
\dfrac{{{a^2}}}{{{b^2}}} = {\left( {\dfrac{a}{b}} \right)^2} = \dfrac{a}{b}.\dfrac{a}{b} = \dfrac{a}{b}.\dfrac{c}{d} = \dfrac{{ac}}{{bd}}\\
\Rightarrow \dfrac{{{a^2} + {c^2}}}{{{b^2} + {d^2}}} = \dfrac{{{a^2}}}{{{b^2}}} = \dfrac{{ac}}{{bd}}
\end{array}\)
