\(\dfrac{3x}{5}=\dfrac{2y}{3}\Leftrightarrow\dfrac{3x}{5}.\dfrac{1}{6}=\dfrac{2y}{3}.\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{3x}{30}=\dfrac{2y}{18}\)
\(\Leftrightarrow\dfrac{x}{10}=\dfrac{y}{9}\)
Suy ra: \(\dfrac{x^2}{100}=\dfrac{y^2}{81}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x^2}{100}=\dfrac{y^2}{81}=\dfrac{x^2-y^2}{100-81}=\dfrac{38}{19}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=2.100=200\\y^2=2.81=162\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\pm\sqrt{200}\\y=\pm\sqrt{162}\end{matrix}\right.\)
\(\dfrac{3x}{5}=\dfrac{2y}{3}\Leftrightarrow\dfrac{3x}{5}.\dfrac{1}{6}=\dfrac{2y}{3}.\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{3x}{30}=\dfrac{2y}{18}\)
\(\Leftrightarrow\dfrac{x}{10}=\dfrac{y}{9}\)
\(\Rightarrow\dfrac{x^2}{100}=\dfrac{y^2}{81}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x^2}{100}=\dfrac{y^2}{81}=\dfrac{x^2-y^2}{100-81}=\dfrac{38}{19}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=2.100=200\\y^2=2.81=162\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm\sqrt{200}\\y=\pm\sqrt{162}\end{matrix}\right.\)
Vậy .........
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