Question

tim x,y biet   (3/5).x = (2/3).y va x² - y² = 38

Answer 1

Đặt \(\frac{3}{5}.x=\frac{2}{3}.y=k\) => \(x=\frac{5}{3}.k;y=\frac{3}{2}.k\)

=> \(x^2-y^2=\left(\frac{5k}{3}\right)^2-\left(\frac{3k}{2}\right)^2=\frac{25}{9}k^2-\frac{9}{4}k^2=\left(\frac{25}{9}-\frac{9}{4}\right)k^2=\frac{19}{36}k^2\)

=> \(\frac{19}{36}k^2=38\)=> k² = 72 => k = \(6\sqrt{2}\) hoặc - \(6\sqrt{2}\)

k = \(6\sqrt{2}\) => x = \(10\sqrt{2}\); y = \(9\sqrt{2}\)

k = - \(6\sqrt{2}\) => x = - \(10\sqrt{2}\); y = - \(9\sqrt{2}\)

Vậy,,,