c: \(10\frac{3}{10}-\left(9,5-0,25\cdot18\right):0,5\)
\(=10,3-\left(9,5-4,5\right):0,5\)
=10,3-5:0,5
=10,3-10=0,3
\(1\frac15-1\frac12\)
=1,2-1,5
=-0,3
\(\frac{10\frac{3}{10}-\left(9,5-0,25\cdot18\right):0,5}{1\frac15-1\frac12}\)
=0,3:(-0,3)
=-1
d: \(\frac{3}{49}\cdot\frac{19}{2}-\frac{3}{49}\cdot\frac52-\left(\frac{1}{20}-\frac14\right)^2\cdot\left(-\frac12-\frac{193}{14}\right)\)
\(=\frac{3}{49}\left(\frac{19}{2}-\frac52\right)-\left(\frac{1}{20}-\frac{5}{20}\right)^2\cdot\frac{-200}{14}\)
\(=\frac{3}{49}\cdot\frac{14}{2}-\left(-\frac{4}{20}\right)^2\cdot\frac{-100}{7}\)
\(=\frac{3}{49}\cdot7-\frac{1}{25}\cdot\frac{-100}{7}=\frac37+\frac47=1\)
