a/ \(\dfrac{x}{5}=\dfrac{y}{3}\Leftrightarrow\dfrac{x^2}{25}=\dfrac{y^2}{9}\)
Áp dụng t,c dãy tỉ số bằng nhau ta có :
\(\dfrac{x^2}{25}=\dfrac{y^2}{9}=\dfrac{x^2-y^2}{25-9}=\dfrac{4}{16}=\dfrac{1}{4}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x^2}{25}=\dfrac{1}{4}\\\dfrac{y^2}{9}=\dfrac{1}{4}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=6,25\\x=-6,25\end{matrix}\right.\\\left[{}\begin{matrix}y=2,25\\y=-2,25\end{matrix}\right.\end{matrix}\right.\)
Vậy -.
b/ \(\dfrac{x}{3}=\dfrac{y}{5}\)
\(\Leftrightarrow\dfrac{x^2}{9}=\dfrac{y^2}{25}\)
\(\Leftrightarrow\dfrac{2x^2}{18}=\dfrac{y^2}{25}\)
Áp dụng t,c dãy tỉ số bằng nhau ta có :
\(\dfrac{2x^2}{18}=\dfrac{y^2}{25}=\dfrac{2x^2-y^2}{18-25}=\dfrac{-28}{-7}=4\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2x^2}{18}=4\\\dfrac{y^2}{25}=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\\\left[{}\begin{matrix}y=10\\y=-10\end{matrix}\right.\end{matrix}\right.\)
Vậy ..
