Question

Tìm x;y khác 0 biết:  \(\frac{x+y}{5}=\frac{x-y}{1}=\frac{x.y}{2}\)

Answer 1

\(\frac{x+y}{5}=\frac{x-y}{1}\)

=>\(\frac{x}{5}+\frac{y}{5}=x-y\)

=>\(\frac{y}{5}+y=x-\frac{x}{5}\)

=>\(\frac{y}{5}+\frac{5y}{5}=\frac{5x}{5}-\frac{x}{5}\)

=>\(\frac{y+5y}{5}=\frac{5x-x}{5}\)

=>\(\frac{6y}{5}=\frac{4x}{5}\)

=>6y=4x

=>\(y=\frac{4}{6}.x\)

Lại có: \(\frac{x-y}{1}=\frac{x.y}{2}\)

=>2.(x-y)=x.y

=>\(2.\left(x-\frac{4}{6}.x\right)=x.y\)

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

=>\(\frac{2}{3}=y\)

=>\(x=\frac{2}{3}:\frac{4}{6}=1\)

Vậy x=1,\(y=\frac{2}{3}\)