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a: \(=\dfrac{2\cdot136-28\cdot6+62\cdot3}{30}\cdot\dfrac{7}{8}=\dfrac{290}{30}\cdot\dfrac{7}{8}=\dfrac{29}{3}\cdot\dfrac{7}{8}=\dfrac{203}{24}\)
b: \(=\dfrac{3}{11}\cdot\dfrac{4}{11}+\dfrac{3}{13}\cdot\dfrac{4}{11}-\dfrac{1}{13}\)
\(=\dfrac{4}{11}\left(\dfrac{3}{11}+\dfrac{3}{13}\right)-\dfrac{1}{13}\)
\(=\dfrac{4}{11}\cdot\dfrac{72}{143}-\dfrac{1}{13}=\dfrac{167}{1573}\)
\(=\dfrac{7}{11}\left(\dfrac{5}{13}+\dfrac{8}{13}\right)-3-\dfrac{7}{11}=-3\)
\(1.\dfrac{-7}{18}+\dfrac{-5}{12}-\dfrac{-13}{18}\text{=}\left(\dfrac{-7}{18}-\dfrac{-13}{18}\right)+\dfrac{-5}{12}\text{=}\dfrac{1}{3}+\dfrac{-5}{12}\text{=}\dfrac{-1}{12}\)
\(2.\dfrac{-13}{17}+\dfrac{-13}{21}+\dfrac{-4}{17}\text{=}\left(\dfrac{-13}{17}+\dfrac{-4}{17}\right)+\dfrac{-13}{21}\text{=}-1+\dfrac{-13}{21}\text{=}\dfrac{-34}{21}\)
\(3.\dfrac{-13}{10}-\dfrac{-4}{13}+\dfrac{-11}{10}\text{=}\dfrac{-12}{5}-\dfrac{-4}{13}\text{=}\dfrac{-136}{65}\)
\(4.\dfrac{13}{17}\times\left(\dfrac{-4}{5}+\dfrac{-3}{4}\right)\text{=}\dfrac{13}{17}\times\dfrac{-31}{20}\text{=}\dfrac{-403}{340}\)
\(5.\left(\dfrac{-5}{12}\times\dfrac{-9}{20}\right)\times\dfrac{-7}{17}\text{=}\dfrac{3}{16}\times\dfrac{-7}{17}\text{=}\dfrac{-21}{272}\)
\(6.\dfrac{11}{23}\times\left(\dfrac{5}{9}+\dfrac{17}{9}-\dfrac{13}{9}\right)\text{=}\dfrac{11}{23}\times1\text{=}\dfrac{11}{23}\)
\(\dfrac{25-\dfrac{1}{11}+\dfrac{4}{13}-\dfrac{4}{15}}{50-\dfrac{2}{11}+\dfrac{8}{13}-\dfrac{8}{15}}\)
\(=\dfrac{25-\dfrac{1}{11}+\dfrac{4}{13}-\dfrac{4}{15}}{2\left(25-\dfrac{1}{11}+\dfrac{4}{13}-\dfrac{4}{15}\right)}=\dfrac{1}{2}\)
\(a.\)
\(\dfrac{2}{9}+\dfrac{-3}{10}+-\dfrac{7}{10}=\dfrac{2}{9}-1=\dfrac{2}{9}-\dfrac{9}{9}=-\dfrac{7}{9}\)
\(b.\)
\(\dfrac{-11}{6}+\dfrac{2}{5}+\dfrac{-1}{6}=\left(-\dfrac{11}{6}+-\dfrac{1}{6}\right)+\dfrac{2}{5}=-2+\dfrac{2}{5}=\dfrac{-10}{5}+\dfrac{2}{5}=\dfrac{-8}{5}\)
\(c.\)
\(-\dfrac{5}{8}+\dfrac{12}{7}+\dfrac{13}{8}+\dfrac{2}{7}=\left(-\dfrac{5}{8}+\dfrac{13}{8}\right)+\left(\dfrac{12}{7}+\dfrac{2}{7}\right)=1+2=3\)
\(\dfrac{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{11}}{\dfrac{13}{4}-\dfrac{13}{5}+\dfrac{13}{7}-\dfrac{13}{11}}\\ =\dfrac{\dfrac{1311}{1540}}{\dfrac{2041}{1540}}\\ =\dfrac{1311}{2041}\)
a, \(4\dfrac{5}{37}\)-\(3\dfrac{4}{5}\)+ \(8\dfrac{15}{29}\)- \(3\dfrac{5}{37}\)+ \(6\dfrac{14}{29}\)
=(\(4\dfrac{5}{37}\)-\(3\dfrac{5}{37}\))+(\(8\dfrac{15}{29}\)+\(6\dfrac{14}{29}\))-\(3\dfrac{4}{5}\)
=(4-3)+(\(\dfrac{5}{37}\)-\(\dfrac{5}{37}\))+(8+6)+(\(\dfrac{15}{29}\)+\(\dfrac{14}{29}\))-3\(\dfrac{4}{5}\)
=1+ 15-\(3\dfrac{4}{5}\)=13-\(\dfrac{4}{5}\)=\(\dfrac{61}{5}\)
b, 60\(\dfrac{7}{13}\)+ 50\(\dfrac{8}{13}\)-11\(\dfrac{2}{13}\)
=(60+50-11)+(\(\dfrac{7}{13}\)+ \(\dfrac{8}{13}\)-\(\dfrac{2}{13}\))
=99+1=100
c, đáp án bằng \(\dfrac{-2}{3}\). bạn tự tính nha
\(B=\dfrac{10}{13}+\dfrac{3}{13}\cdot\dfrac{3}{11}+\dfrac{3}{13}\cdot\dfrac{8}{11}\)
\(B=\left(\dfrac{3}{13}+\dfrac{10}{13}\right)\cdot\left(\dfrac{3}{11}+\dfrac{8}{11}\right)\)
\(B=1\cdot1\)
\(B=1\)