\(\frac{a}{5}=\frac{b}{3},\frac{b}{7}=\frac{c}{9}\Rightarrow\frac{a}{35}=\frac{b}{21},\frac{b}{21}=\frac{c}{27}\Rightarrow\frac{a}{35}=\frac{b}{21}=\frac{c}{27}\)
\(\Rightarrow\frac{a}{35}=\frac{b}{21}=\frac{c}{27}=\frac{a+b}{35+21}=\frac{a+b}{56}=\frac{b-c}{21-27}=\frac{b-c}{-6}\)(T/C)
\(\Rightarrow\frac{a+b}{56}=\frac{b-c}{-6}=\frac{a+b}{b-c}=\frac{56}{-6}=-\frac{28}{3}\)
Giải:
Ta có: \(a:b=5:3\Rightarrow\frac{a}{5}=\frac{b}{3}\Rightarrow\frac{a}{35}=\frac{b}{21}\)
\(b:c=7:9\Rightarrow\frac{b}{7}=\frac{c}{9}\Rightarrow\frac{b}{21}=\frac{c}{27}\)
\(\Rightarrow\frac{a}{35}=\frac{b}{21}=\frac{c}{27}\)
Đặt \(\frac{a}{35}=\frac{b}{21}=\frac{c}{27}=k\)
\(\Rightarrow a=35k,b=21k,c=27k\)
Từ đó \(\frac{a+b}{b-c}=\frac{35k+21k}{21k-27k}=\frac{56k}{-6k}=\frac{-28}{3}\)
Vậy \(\frac{a+b}{b-c}=\frac{-28}{3}\)
