a) [tex]3 \frac{6}{9} -( \frac{3}{9} - \frac{15}{90})-(2 \frac{75}{100} -2 \frac{3}{9})= [/tex]
[tex]3 \frac{2}{3} - \frac{45}{90} -( \frac{3}{4}- \frac{1}{3})= \frac{11}{3} - \frac{1}{2}- \frac{5}{3}= [/tex][tex] \frac{6}{3}- \frac{1}{2}= \frac{3}{2} [/tex]
b) [tex] \frac{3(1+2+...+100)}{2(1+2+...+100)} - \frac{1+2+...+100}{4(1+2+...+100)}= \frac{3}{2}- \frac{1}{4} = \frac{5}{4} [/tex]
