a, \(\frac{3.7-3.5}{5+12}=\frac{3.\left(7-5\right)}{17}\)
\(=\frac{3.2}{17}\)
\(=\frac{6}{17}\)
b, \(\frac{9.6-9.3}{27}=\frac{9.\left(6-3\right)}{27}\)
\(=\frac{9.3}{27}\)
\(=\frac{27}{27}=1\)
nha
\(\frac{3.7-3.5}{6+12}\)=\(\frac{3.\left(7-5\right)}{3.\left(2+4\right)}\)=\(\frac{2}{6}\)=\(\frac{1}{2}\)
\(\frac{9.6-9.3}{27}=\)\(\frac{9.3}{9.3}\)=1