Từ \(\frac{x}{5}=\frac{y}{4}\Rightarrow\frac{x^2}{25}=\frac{y^2}{16}\)
Áp dụng tính chất dãy tit số bằng nhau ta có:
\(\frac{x}{5}=\frac{y}{4}=\frac{x^2}{25}=\frac{y^2}{16}=\frac{x^2-y^2}{25-16}=\frac{4}{9}\)
\(\Rightarrow\left\{\begin{matrix}\frac{x^2}{25}=\frac{4}{9}\Rightarrow x^2=\frac{4\cdot25}{9}=\frac{100}{9}\Rightarrow x=\pm\frac{10}{3}\\\frac{y^2}{16}=\frac{4}{9}\Rightarrow y^2=\frac{4\cdot16}{9}=\frac{64}{9}\Rightarrow y=\pm\frac{8}{3}\end{matrix}\right.\)
Ta có: \(\frac{x}{5}\) = \(\frac{y}{4}\) => \(\frac{xy}{5.4}\) = \(\frac{x^2}{25}\) = \(\frac{y^2}{16}\)
Áp dụng tc dãy tỉ số bằng nhau ta có:
\(\frac{x^2}{25}\) = \(\frac{y^2}{16}\) = \(\frac{x^2-y^2}{25-16}\) = \(\frac{4}{9}\)
Do \(\frac{x^2}{25}\) = \(\frac{4}{9}\) => x2 = \(\frac{100}{9}\) => x = \(\sqrt{\frac{100}{9}}\)
\(\frac{y^2}{16}\) = \(\frac{4}{9}\) => y2 = \(\frac{64}{9}\) => y = \(\sqrt{\frac{64}{9}}\)
Vậy x = \(\sqrt{\frac{100}{9}}\) và y = \(\sqrt{\frac{64}{9}}\).