Odpowiedź:
[tex]c) (\frac{1}{2}m-\frac{1}{5}):2 + (\frac{1}{2} m+\frac{2}{5} ):4 = \frac{5m-2}{10}*\frac{1}{2}+\frac{3m+4}{6}*\frac{1}{4} = \frac{1}{4}*(\frac{5m-2}{10}*2+\frac{3m+4}{6}) = \frac{1}{4}*(\frac{5m-2}{5} + \frac{3m+4}{6} )[/tex]
[tex]d) (\frac{2}{3}a+\frac{3}{4}b):6-(\frac{1}{3}a+\frac{1}{2}b):4 = \frac{8a+9b}{12}*\frac{1}{6}-\frac{\frac{1}{3}a+\frac{1}{2}b }{4} = \frac{8a+9b}{72}-\frac{\frac{2a+3b}{6} }{4} = \frac{8a+9b}{72}-\frac{2a+3b}{24} = \frac{8a+9b-3(2a+3b)}{72} = \frac{8a+9b-6a-9b}{72} = \frac{8a-6a}{72} = \frac{2a}{72}[/tex]
[tex]e) 6(\frac{1}{6}u+\frac{1}{12} )-(\frac{1}{2}u+\frac{3}{4} ):3 = u+\frac{1}{2}-\frac{\frac{1}{2}u+\frac{3}{4} }{3} = u+\frac{1}{2}-\frac{\frac{2u+3}{4} }{3} = u+\frac{1}{2}-\frac{2u+3}{12} = \frac{12u+6-(2u+3)}{12} = \frac{12u+6-2u-3}{12} = \frac{10u+3}{12}[/tex]
