Răspuns :
a) [tex]( \frac{1}{7}+ \frac{7}{11}- \frac{2010}{2011})+( \frac{6}{7}+ \frac{4}{11}- \frac{2010}{2011})=(- \frac{34110}{154847})+ \frac{34264}{154847}[/tex][tex]=\frac{34110}{154847}+ \frac{34264}{154847} = \frac{-34110+34264}{154847}= \frac{154}{154847}= \frac{2}{2011} [/tex] ≈ 0.00099453
b) [tex]( \frac{5}{3}+ \frac{7}{5}+ \frac{9}{7})-( \frac{2}{3}+ \frac{2}{5}+ \frac{2}{7})= \frac{457}{105}- \frac{142}{105}= \frac{457-142}{105}= \frac{315}{105}=3 [/tex]
b) [tex]( \frac{5}{3}+ \frac{7}{5}+ \frac{9}{7})-( \frac{2}{3}+ \frac{2}{5}+ \frac{2}{7})= \frac{457}{105}- \frac{142}{105}= \frac{457-142}{105}= \frac{315}{105}=3 [/tex]