Question

Полное решение с объяснением ​

Answer 1

$\displaystyle 1)\frac{1}{8}(6a-4)-\frac{1}{12}(10b+2)=\frac{1}{8}*2(3a-2)-\frac{1}{12}*2(5b+1)=\frac{1}{4}(3a-2)-\frac{1}{6}(5b+1)=\\ \\ \frac{3}{4}a-\frac{1}{2}-\frac{5}{6}b-\frac{1}{6}=\frac{3}{4}a-\frac{2}{3}-\frac{5}{6}b\\\\ 2)\frac{1}{8}(6*1\frac{1}{3}-4)-\frac{1}{12}(10*\frac{3}{5}+2)=\frac{1}{8}(6*\frac{4}{3}-4)-\frac{1}{12}(2*3+2)=\frac{1}{8}(2*4-4)-\\ \\ \frac{1}{12}(6+2)=\frac{1}{8}(8-4)-\frac{1}{12}*8=\frac{1}{8}*4-\frac{1}{3}*2=\frac{1}{2}-\frac{2}3}=-\frac{1}{6}$