\(\dfrac{a+3}{5}=\dfrac{b-2}{3}=\dfrac{c-1}{7}\)
\(\Leftrightarrow\dfrac{3a+9}{15}=\dfrac{5b-10}{15}=\dfrac{7c-7}{49}\)
Theo tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{3a+9}{15}=\dfrac{5b-10}{15}=\dfrac{7c-7}{49}=\dfrac{3a+9-\left(5b-10\right)+\left(7c-7\right)}{15-15+49}=\dfrac{\left(3a-5b+7c\right)+9+10-7}{49}=\dfrac{86+12}{49}=\dfrac{98}{49}=2\)
\(\Rightarrow\left\{{}\begin{matrix}3a+9=30\\5b-10=30\\7c-7=98\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3a=21\\5b=40\\7c=105\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=7\\b=8\\c=15\end{matrix}\right.\)