Question

tim x ,y 

\(\frac{x}{5}=\frac{y}{3}va\)\(^{x^2-y^2=4}\)

 

Answer 1

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

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

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

=>\(x^2=25.\frac{1}{4}=\frac{25}{4}=>x=\frac{5}{2},-\frac{5}{2}\)

Xét \(x=\frac{5}{2}\)=>\(\frac{x}{5}=\frac{y}{3}=\frac{\frac{5}{2}}{5}=\frac{1}{2}=>y=\frac{1}{2}.3=\frac{3}{2}\)

Xét \(x=-\frac{5}{2}\)=>\(\frac{x}{5}=\frac{y}{3}=\frac{-\frac{5}{2}}{5}=-\frac{1}{2}=>y=-\frac{1}{2}.3=-\frac{3}{2}\)