Ta có \(\frac{a+3}{5}=\frac{b-2}{3}=\frac{c-1}{7}\)\(=\frac{3a+9}{15}=\frac{5b-10}{15}=\frac{7c-7}{49}=\frac{3a+9-5b+10+7c-7}{15-15+49}\)\(=\frac{\left(3a-5b+7c\right)+\left(9+10-7\right)}{49}=\frac{86+12}{49}=\frac{98}{49}=2\)
\(=>\hept{\begin{cases}a+3=10\\b-2=6\\c-1=14\end{cases}\left(=\right)\hept{\begin{cases}a=7\\b=8\\c=15\end{cases}}}\)
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