Vă rog mult promit 50 puncte și o coroana ex 7

[tex]7) a)~(1+ \frac{1}{2}):(1- \frac{1}{2}) = (\frac{2+1}{2}):( \frac{2-1}{2}) = \frac{3}{2}: \frac{1}{2} = \frac{3}{2}*2 = 3 \\ \\ b) \frac{1}{2}:( \frac{2}{5} - \frac{1}{10}) = \frac{1}{2}:( \frac{2*2-1}{10}) = \frac{1}{2}: \frac{3}{10} = \frac{1}{2}* \frac{10}{3} = \frac{5}{3} \\ \\ c) ( \frac{7}{36} - \frac{1}{24} ): \frac{7}{6}= \frac{7*2-1*3}{72}: \frac{7*12}{72}= \frac{11}{72}* \frac{72}{84}= \frac{11}{84}[/tex]
[tex]d)( \frac{1}{4}- \frac{1}{8}): \frac{1}{12}= \frac{6-3}{24} : \frac{2}{24}= \frac{3}{24}* \frac{24}{2} = \frac{3}{2} \\ \\ e)2:(3 \frac{3}{7}+ \frac{1}{4})=2:( \frac{3*7+3}{7}+ \frac{1}{4})=2:( \frac{24}{7}+ \frac{1}{4})=2: \frac{24*4+7}{28}=2* \frac{28}{103} \\ = \frac{56}{103} \\ \\ f)3:(3- \frac{2}{3})=3: \frac{3*3-2}{3}=3* \frac{3}{7}= \frac{9}{7} [/tex]
[tex]g) \frac{24}{10}: \frac{4}{5}+3(1 \frac{1}{2}+ \frac{1}{2})= \frac{24}{10}* \frac{5}{4}+3* \frac{1*2+1+1}{2}= 3+3* \frac{4}{2}=3+3*2=8 \\ \\ h)1- \frac{2}{5}*1 \frac{1}{4}+3:1 \frac{1}{5}=1- \frac{2}{5}* \frac{1*4+1}{4}+3: \frac{1*5+1}{5}=1- \frac{2}{5}* \frac{5}{4}+3* \frac{5}{6}= \\ \\1- \frac{1}{2}+ \frac{5}{2}= \frac{2-1+5}{2}= \frac{6}{2}=3 \\ \\ i)( \frac{3}{10}- \frac{3}{20})*1\frac{1}{3}= \frac{3*2-3}{20}* \frac{1*3+1}{3}= \frac{3}{20}* \frac{4}{3}= \frac{1}{5}[/tex]
[tex]j) \frac{1}{2}+ \frac{1}{4}: \frac{1}{3}: \frac{3}{4}= \frac{1}{2}+ \frac{1}{4}*3* \frac{4}{3}= \frac{1}{2}+1= \frac{1+2}{2}= \frac{3}{2} \\ \\ k) \frac{36}{59}*( \frac{1}{45}+ \frac{7}{60}+ \frac{1}{40})= \frac{36}{59}* \frac{8+6+9}{360}= \frac{36}{59}* \frac{23}{360}= \frac{23}{590} \\ \\ l)( \frac{3}{35}+ \frac{1}{70}+ \frac{1}{105}): \frac{1}{140}= \frac{3*6+3+2}{210}*140= \frac{23}{210}*140= \frac{46}{3}[/tex]