Theo de bai ta co
\(\frac{a+b}{6}=\frac{b+c}{7}=\frac{c+a}{8},a+b+c=14\)
Ap dung tinh chat day ti so bang nhau ta co
=\(\frac{a+b+b+c+c+a}{6+7+8}=\frac{2\left(a+b+c\right)}{21}=\frac{2.14}{21}=\frac{4}{3}\)
=> \(\frac{a+b}{6}=\frac{4}{3}\)=>a+b=\(6.\frac{4}{3}\)=8
=>b=8-a
=>\(\frac{b+c}{7}=\frac{4}{3}\) => b+c=\(7.\frac{4}{3}\)=\(\frac{28}{3}\)
=>b=\(\frac{28}{3}-c\)
=> \(\frac{c+a}{8}=\frac{4}{3}\)=>c+a=\(8.\frac{4}{3}\)=\(\frac{32}{3}\)
=>c=\(\frac{32}{3}-a\)
Ta co b=8-a ; b=\(\frac{28}{3}-c\)
=>8-a=\(\frac{28}{3}-c\)
-a=\(\frac{28}{3}-8\)-c
-a=\(\frac{4}{3}-c\)
c=\(\frac{4}{3}+a\)
Ma ta co c=\(\frac{32}{3}-a\)
=>\(\frac{4}{3}+a=\frac{32}{3}-a\)
\(2a=\frac{32}{3}-\frac{4}{3}=\frac{28}{3}\)
=>a=\(\frac{28}{3}:2=\frac{14}{3}\)
Ta co c=\(\frac{32}{3}-a\)
=>c=\(\frac{32}{3}-\frac{14}{3}=6\)
