Question

Va rog foarte frumos, repedee!!!

Answer 1

[tex]\it\ a)\ \dfrac{^{3)}1}{\ 2}+\dfrac{^{2)}2}{\ 3} = \dfrac{3}{6}+\dfrac{4}{6} =\dfrac{3+4}{6} =\dfrac{7}{6} =1\dfrac{1}{6} \\\;\\ \\\;\\ b)\ \dfrac{^{3)}3}{\ 4}+\dfrac{^{2)}1}{\ 6} =\dfrac{9}{12}+\dfrac{2}{12} = \dfrac{9+2}{12} = \dfrac{11}{12} [/tex]


[tex]\it c)\ \dfrac{^{3)}1}{\ \ 20}-\dfrac{^{2)}1}{\ \ 30} = \dfrac{3}{60} - \dfrac{2}{60} =\dfrac{3-2}{60} =\dfrac{1}{60} \\\;\\ \\\;\\ d)\ \dfrac{^{3)}5}{12}-\dfrac{^{2)}7}{18} = \dfrac{15}{36}-\dfrac{14}{36} =\dfrac{15-14}{36} =\dfrac{1}{36} [/tex]



[tex]\it\ e)\ \dfrac{^{3)}1}{50} +\dfrac{^{2)}1}{75} =\dfrac{3}{150}+\dfrac{2}{150} =\dfrac{3+2}{150} = \dfrac{\ 5^{(5}}{150} = \dfrac{1}{30} \\\;\\ \\\;\\ f)\ \dfrac{^{3)}1}{2}+\dfrac{^{2)}2}{3}+\dfrac{5}{6}= \dfrac{3}{6} +\dfrac{4}{6}+\dfrac{5}{6} =\dfrac{3+4+5}{6} =\dfrac{12}{6} =2[/tex]


[tex]\it \ell) \ ^{6)} 2+\dfrac{^{3)}1}{2} -\dfrac{^{2)}4}{3}-\dfrac{1}{6} = \dfrac{12}{6} + \dfrac{3}{6}- \dfrac{8}{6}- \dfrac{1}{6}=\dfrac{12+3-8-1}{6}= \\\;\\ \\\;\\ =\dfrac{6}{6} =1[/tex]