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\(A=\dfrac{1}{3}\cdot\dfrac{4}{5}+\dfrac{1}{3}\cdot\dfrac{6}{5}+\dfrac{2}{3}\\ =\dfrac{1}{3}\cdot\left(\dfrac{4}{5}+\dfrac{6}{5}\right)+\dfrac{2}{3}\\ =\dfrac{1\cdot2}{3}+\dfrac{2}{3}\\ =\dfrac{2}{3}+\dfrac{2}{3}\\ =\dfrac{4}{3}\)
B =\(\dfrac{-5}{6}.\dfrac{4}{19}+\dfrac{-7}{12}.\dfrac{4}{19}\)-\(\dfrac{40}{57}\)
=\(\dfrac{4}{19}.\left[\dfrac{-5}{6}+\dfrac{-7}{12}\right]-\dfrac{40}{57}\)
=\(\dfrac{4}{19}.\left[\dfrac{-10}{12}+\dfrac{-7}{12}\right]-\dfrac{40}{57}\)
=\(\dfrac{4}{19}.\dfrac{-17}{12}-\dfrac{40}{57}\)
=\(\dfrac{-17}{57}\)-\(\dfrac{40}{57}\)
=\(\dfrac{-17}{57}+\dfrac{-40}{57}\)
=\(\dfrac{-57}{57}=-1\)
\(D=\dfrac{1}{2}+\dfrac{-1}{5}+\dfrac{-5}{7}+\dfrac{1}{6}+\dfrac{-3}{35}+\dfrac{1}{3}+\dfrac{1}{41}\)
\(D=\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{3}\right)+\left(\dfrac{-1}{5}+\dfrac{-5}{7}+\dfrac{-3}{35}\right)+\dfrac{1}{41}\)
\(D=1+-1+\dfrac{1}{41}\)
\(D=0+\dfrac{1}{41}\)
\(D=\dfrac{1}{41}\)
\(C=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)+\left(\dfrac{-3}{4}+\dfrac{-1}{36}+\dfrac{-2}{9}\right)+\dfrac{1}{57}\)
\(=\dfrac{5+9+1}{15}+\dfrac{-27-1-8}{36}+\dfrac{1}{57}\)
=1/57
\(E=\left(-\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\left(\dfrac{3}{5}+\dfrac{4}{35}+\dfrac{2}{7}\right)+\dfrac{1}{127}=\dfrac{1}{127}\)
ta có
\(2.\left(\dfrac{1}{3}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{6}\right)\) \(5.\left(\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{1}{6}+\dfrac{1}{11}\right)\)
_______________________ X ________________________
\(4.\left(\dfrac{1}{3}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{6}\right)\) \(9.\left(\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{1}{6}\dfrac{1}{11}\right)\)
= \(\dfrac{2}{4}X\dfrac{5}{9}\)= \(\dfrac{10}{36}\)= \(\dfrac{5}{18}\)
a: \(=\dfrac{-28}{36}+\dfrac{15}{36}-\dfrac{26}{36}=\dfrac{-39}{36}=\dfrac{-13}{12}\)
b: \(=\dfrac{11}{9}\left(\dfrac{15}{4}-\dfrac{7}{4}-\dfrac{5}{4}\right)=\dfrac{11}{9}\cdot\dfrac{3}{4}=\dfrac{11}{12}\)
c: \(=15+\dfrac{9}{7}+6+\dfrac{2}{3}-5-\dfrac{5}{9}\)
\(=16+\dfrac{88}{63}=\dfrac{1096}{63}\)
d: \(=\dfrac{5}{6}-\dfrac{1}{3}+\dfrac{2}{18}\)
\(=\dfrac{15-6+2}{18}=\dfrac{11}{18}\)
c,
= \(\dfrac{5}{9}.\left(\dfrac{7}{13}+\dfrac{9}{13}+\dfrac{-3}{13}\right)\)
= \(\dfrac{5}{9}.1\)
= \(\dfrac{5}{9}\)
Trời ơi cái đề bài !!!
Thoy thì làm từng câu vậy
a) \(I=10101.\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{111111}\right)\)
\(I=10101.\left(\dfrac{10}{222222}+\dfrac{5}{222222}-\dfrac{8}{222222}\right)\)
\(I=10101.\left(\dfrac{15}{222222}-\dfrac{8}{222222}\right)\)
\(I=10101.\dfrac{7}{222222}\)
\(I=\dfrac{7}{22}\)
\(-2\dfrac{1}{4}.\)\(\left(3\dfrac{5}{12}-1\dfrac{2}{9}\right)\)
=\(\dfrac{-9}{4}\).\(\left(\dfrac{41}{12}-\dfrac{11}{9}\right)\)
=\(\dfrac{-9}{4}.\dfrac{41}{12}-\dfrac{-9}{4}.\dfrac{11}{9}\)
=\(\dfrac{-123}{16}-\dfrac{-11}{4}\)
=\(\dfrac{-123}{16}-\dfrac{-44}{16}\)
=\(\dfrac{-79}{16}\)
\(\left(-25\%+0,75+\dfrac{7}{12}\right)\div\left(-2\dfrac{1}{8}\right)\)
=\(\left(\dfrac{-1}{4}+\dfrac{3}{4}+\dfrac{7}{12}\right)\div\left(\dfrac{-17}{8}\right)\)
=\(\left(\dfrac{-3}{12}+\dfrac{9}{12}+\dfrac{7}{12}\right).\dfrac{-8}{17}\)
=\(\dfrac{13}{12}.\dfrac{-8}{17}=\dfrac{-26}{51}\)
\(A=\dfrac{6}{7}+\dfrac{1}{7}.\dfrac{2}{7}+\dfrac{1}{7}.\dfrac{5}{7}.\)
\(A=\dfrac{6}{7}+\dfrac{1}{7}\left(\dfrac{2}{7}+\dfrac{5}{7}\right).\)
\(A=\dfrac{6}{7}+\dfrac{1}{7}.1.\)
\(A=\dfrac{6}{7}+\dfrac{1}{7}=1.\)
Vậy \(A=1.\)
\(B=\dfrac{40}{9}.\dfrac{13}{3}-\dfrac{4}{3}.\dfrac{40}{9}.\)
\(B=\dfrac{4}{9}.\dfrac{13}{3}-\dfrac{4}{9}.\dfrac{40}{3}.\)
\(B=\dfrac{4}{9}\left(\dfrac{13}{3}-\dfrac{40}{3}\right).\)
\(B=\dfrac{4}{9}.\left(-9\right).\)
\(B=-4.\)
Vậy \(B=-4.\)
\(a)\left(2\dfrac{5}{6}+1\dfrac{4}{9}\right):\left(10\dfrac{1}{12}-9\dfrac{1}{2}\right)\)
\(=\left(\dfrac{17}{6}+\dfrac{13}{9}\right):\left(10\dfrac{1}{12}-9\dfrac{6}{12}\right)\)
\(=\left(\dfrac{153}{54}+\dfrac{78}{54}\right):\left(1\dfrac{-5}{12}\right)\)
\(=\dfrac{231}{54}:\dfrac{7}{12}\)
\(=\dfrac{198}{27}\)
\(b)\dfrac{0,8\left(\dfrac{4}{5}:1,25\right)}{0,64-\dfrac{1}{25}}\)
\(=\dfrac{0,8\left(0,8:1,25\right)}{0,64-0,04}\)
\(=\dfrac{0,8.0,64}{0,6}\)
\(=\dfrac{0,512}{0,6}\)\(=\dfrac{64}{75}\)
\(\dfrac{-5}{6}.\dfrac{4}{9}+\dfrac{-7}{12}.\dfrac{4}{9}-\dfrac{40}{57} \)
\(=\dfrac{4}{9}.\left(\dfrac{-5}{6}+\dfrac{-7}{12}\right)-\dfrac{40}{57}\)
\(=\dfrac{4}{9}.\dfrac{-17}{12}-\dfrac{40}{57}\)
\(=\dfrac{-17}{27}-\dfrac{40}{57}\)
\(=\dfrac{-683}{513}\)
@dangthuylinh
\(\dfrac{-5}{6}.\dfrac{4}{9}+\dfrac{-7}{12}.\dfrac{4}{9}-\dfrac{40}{57}\)
\(=\dfrac{4}{9}.\left(\dfrac{-5}{6}+\dfrac{-7}{12}\right)-\dfrac{40}{57}\)
\(=\dfrac{4}{9}.\dfrac{-17}{12}-\dfrac{40}{57}\)
\(=\dfrac{-17}{27}-\dfrac{40}{57}\)
\(=\dfrac{-683}{513}\)