\(\frac{a+3-3}{5-3}=\frac{b-2+2}{3+2}=\frac{c-1+1}{7+1}=\frac{a}{2}=\frac{b}{5}=\frac{c}{8}=\frac{3a}{6}=\frac{5b}{25}=\frac{7c}{56}=\frac{86}{37}\)

\(\Rightarrow\frac{a}{2}=\frac{86}{37}\Rightarrow a=\frac{172}{37};\frac{b}{5}=\frac{86}{37}\Rightarrow\frac{430}{37};\frac{c}{8}=\frac{86}{37}\Rightarrow\frac{688}{37}\)

\(\dfrac{a+3}{5}=\dfrac{b-2}{3}=\dfrac{c-1}{7}\)

\(\Leftrightarrow\dfrac{3a+9}{15}=\dfrac{5b-10}{15}=\dfrac{7c-7}{49}\)

Theo tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{3a+9}{15}=\dfrac{5b-10}{15}=\dfrac{7c-7}{49}=\dfrac{3a+9-\left(5b-10\right)+\left(7c-7\right)}{15-15+49}=\dfrac{\left(3a-5b+7c\right)+9+10-7}{49}=\dfrac{86+12}{49}=\dfrac{98}{49}=2\)

\(\Rightarrow\left\{{}\begin{matrix}3a+9=30\\5b-10=30\\7c-7=98\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3a=21\\5b=40\\7c=105\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=7\\b=8\\c=15\end{matrix}\right.\)

\(3\left(a+3\right)=5\left(b-2\right) \)
\(3a+9=5b-10\)

\(3a-5b=-9-10=-19\)
\(3a-5b=-19\)

thay -19 vào 3a -5b ta có: -19+7c =86  suy ra c= (86+19)/7=15
thay 15 vào c ta có \(\frac{14}{7}=\frac{a+3}{5}=\frac{b-2}{3}\)hay \(2=\frac{a+3}{5}=\frac{b-2}{3}\)vậy a= 7, b=8 và c=15
Còn một cách khác mà làm biếng tí

Bạn kia làm đúng rồi

Ta có \(\frac{a+3}{5}=\frac{b-2}{3}=\frac{c-1}{7}\)\(=\frac{3a+9}{15}=\frac{5b-10}{15}=\frac{7c-7}{49}=\frac{3a+9-5b+10+7c-7}{15-15+49}\)\(=\frac{\left(3a-5b+7c\right)+\left(9+10-7\right)}{49}=\frac{86+12}{49}=\frac{98}{49}=2\)

\(=>\hept{\begin{cases}a+3=10\\b-2=6\\c-1=14\end{cases}\left(=\right)\hept{\begin{cases}a=7\\b=8\\c=15\end{cases}}}\)

học tốt

\(a+\frac{3}{5}=\frac{b-2}{3}=c-\frac{1}{7}\)

\(a+\frac{3}{5}=b-\frac{2}{3},b-\frac{2}{3}=c-\frac{1}{7}\)

\(b=\frac{15r+19}{15}\)

\(c=\frac{35r+26}{35}\)\(;r\in R\)

ko pic 

lỗi