Va rog ajutati ma dau coroana 100%

a) [tex]2:( \frac{2}{3})^2+ \frac{11}{5}: \frac{11}{10}=2: \frac{4}{9}+ \frac{11}{5}* \frac{10}{11}=2* \frac{4}{9}+2= \frac{9}{2}+2= \frac{9+4}{2}= \frac{13}{4} [/tex]
b) [tex](3- \frac{1}{3})^2+(2- \frac{1}{2})^2-(1 \frac{1}{2})^2=( \frac{8}{3})^2+( \frac{3}{2})^2-( \frac{3}{2})^2= \frac{64}{9}+ \frac{9}{4}- \frac{9}{4}= [/tex][tex] \frac{337}{36}- \frac{9}{4}= \frac{337-81}{36}= \frac{256}{36}= \frac{64}{9} [/tex]
c) [tex][( \frac{1}{2}+ \frac{2}{3}+\frac{5}{6}): \frac{7}{12}-\frac{2}{5} *\frac{5}{6}]*\frac{1}{4}:(5*\frac{12}{25}-1*\frac{1}{3})=[/tex][tex](2*\frac{12}{7}-\frac{1}{3})*\frac{1}{4}:(\frac{12}{5}-\frac{1}{3})=( \frac{24}{7}- \frac{1}{3})* \frac{1}{4}: \frac{31}{15}= \frac{65}{21}* \frac{1}{4}* \frac{15}{31}= [/tex][tex] \frac{65}{7}* \frac{1}{4}* \frac{5}{31}= \frac{65*1*5}{7*4*31}= \frac{325}{868} [/tex]
d) [tex](1- \frac{1}{2})*(1- \frac{1}{3})*(1- \frac{1}{4})*(1- \frac{1}{50})= \frac{1}{2}* \frac{2}{3}* \frac{3}{4} * \frac{49}{50} = \frac{1}{4}* \frac{49}{50}= \frac{1*49}{4*50}= [/tex][tex] \frac{49}{200} [/tex]