Răspuns:
a) [[tex][\frac{3}{4}+\frac{2}{3}:(\frac{3}{4}-\frac{1}{2})].\frac{6}{41}=[\frac{3}{4}+\frac{2}{3}:(\frac{3}{4}-\frac{2}{4})].\frac{6}{41}=[\frac{3}{4}+\frac{2}{3}:\frac{1}{4}].\frac{6}{41}=[\frac{3}{4}+\frac{2}{3}.\frac{4}{1}].\frac{6}{41}=[\frac{3}{4}+\frac{8}{3}].\frac{6}{41}=[\frac{9}{12}+\frac{32}{12}].\frac{6}{41}=\frac{41}{12}.\frac{6}{41}= \frac{1}{2}.\frac{1}{1}=\frac{1}{2}[/tex]
b)[tex][\frac{7}{5}:\frac{14}{15}+\frac{3}{2}:(\frac{9}{4}-1\frac{1}{2})].\frac{2}{7}= [\frac{7}{5}.\frac{15}{14}+\frac{3}{2}:(\frac{9}{4}-\frac{3}{2})].\frac{2}{7}=[\frac{1}{1}.\frac{3}{2}+\frac{3}{2}:(\frac{9}{4}-\frac{6}{4})].\frac{2}{7}=[\frac{3}{2}+\frac{3}{2}:\frac{3}{4}].\frac{2}{7}=[\frac{3}{2}+\frac{3}{2}.\frac{4}{3}].\frac{2}{7}=[\frac{3}{2}+\frac{2}{1}].\frac{2}{7}=[\frac{3}{2}+\frac{4}{2}].\frac{2}{7}=\frac{7}{2}.\frac{2}{7}=\frac{1}{1}=1[/tex]
c)[tex][5\frac{2}{3}+1,(5)-2]:15\frac{2}{3}=[\frac{17}{3}+1\frac{5}{9} -2]:\frac{47}{3}=[\frac{17}{3}+\frac{14}{9} -\frac{2}{1}]:\frac{47}{3}=[\frac{51}{9}+\frac{14}{9} -\frac{18}{9}]:\frac{47}{3}=[\frac{65}{9}-\frac{18}{9}]:\frac{47}{3}=\frac{47}{9}.\frac{3}{47}=\frac{1}{3}.\frac{1}{1}=\frac{1}{3}[/tex]
d)[tex][4+6\frac{2}{3}-2\frac{1}{4} .(\frac{2}{3}-\frac{2}{9})]:4\frac{5}{6}+\frac{29}{4}= [\frac{4}{1} +\frac{20}{3}-\frac{9}{4} .(\frac{6}{9}-\frac{2}{9})]:\frac{29}{6}+\frac{29}{4}=[\frac{12}{3}+\frac{20}{3}-\frac{9}{4}.\frac{4}{9}]:\frac{29}{6}+\frac{29}{4}=[\frac{32}{3}-\frac{9}{4}.\frac{4}{9}]:\frac{29}{6}+\frac{29}{4}=[\frac{32}{3}-\frac{1}{1} ]:\frac{29}{6}+\frac{29}{4}=[\frac{32}{3}-\frac{3}{3}] :\frac{29}{6}+\frac{29}{4}=\\\frac{29}{3}.\frac{6}{29}+ \frac{29}{4}=\frac{1}{1}.\frac{2}{1}+\frac{29}{4}=\\[/tex]
[tex]\frac{2}{1}+\frac{29}{4} =\frac{8}{4}+\frac{29}{4}=\frac{37}{4}[/tex]
Explicație pas cu pas:
