Question

\(\dfrac{x}{3}=\dfrac{y}{5}v\text{à }x^2-y^2=-4\)

Answer 1

Từ \(\dfrac{x}{3}=\dfrac{y}{5}\)=> \(\dfrac{x^2}{9}=\dfrac{y^2}{25}\)

Áp dụng t/c dãy tỉ số bằng nhau, ta có:

\(\dfrac{x^2}{9}=\dfrac{y^2}{25}=\dfrac{x^2-y^2}{9-25}=\dfrac{-4}{-16}=\dfrac{1}{4}\)

=> \(\left\{{}\begin{matrix}\dfrac{x^2}{9}=\dfrac{1}{4}\\\dfrac{y^2}{25}=\dfrac{1}{4}\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x^2=\dfrac{9}{4}\\y^2=\dfrac{25}{4}\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}y=\dfrac{5}{2}\\y=-\dfrac{5}{2}\end{matrix}\right.\end{matrix}\right.\)

Answer 2