\(\frac{2a+7}{5}=\frac{3b-3}{4}=\frac{c+5}{3}\)
=> \(\frac{4a+14}{10}=\frac{12b-12}{16}=\frac{3c+15}{9}=\frac{4a+14+12b-12-3c-15}{10+16-9}\)
\(=\frac{\left(4a+12b-3c\right)-13}{17}=\frac{64-13}{17}=3\)
=> \(\hept{\begin{cases}2a+7=15\\3b-3=12\\c+5=9\end{cases}}\Rightarrow\hept{\begin{cases}a=4\\b=5\\c=4\end{cases}}\)
Vậy a = 4 ; b = 5 ; c = 4