Question

Biet \(\frac{x}{y}=\frac{4}{3}\). Khi do \(\frac{x-y}{x+y}=\)?

Answer 1

\(\frac{x-y}{x+y}=\frac{4-3}{4+3}=\frac{1}{7}\)

Answer 2

\(\frac{x}{y}=\frac{4}{3}=>\frac{x}{4}=\frac{y}{3}=\frac{x-y}{4-3}=\frac{x+y}{4+3}=\frac{x-y}{1}=\frac{x+y}{7}\)

=>\(\frac{x-y}{1}=\frac{x+y}{7}=>\frac{x-y}{x+y}=\frac{1}{7}\)

Answer 3

\(\frac{x}{y}=\frac{4}{3}\Leftrightarrow\frac{x}{4}=\frac{y}{3}=\frac{x-y}{4-3}=\frac{x+y}{4+3}\Leftrightarrow\frac{x-y}{x+y}=\frac{4-3}{4+3}=\frac{1}{7}\)