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\(=\left(\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}+\dfrac{3\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}\right)\cdot\dfrac{2005}{1890}+115\)
\(=\left(\dfrac{3}{5}-\dfrac{3}{5}\right)\cdot\dfrac{2005}{1890}+115\)
=115
a) \(\dfrac{3}{4}-\dfrac{1}{8}=\dfrac{6}{8}-\dfrac{1}{8}=\dfrac{6-1}{8}=\dfrac{5}{8}\)
b) \(\dfrac{2}{6}-\dfrac{5}{18}=\dfrac{6}{18}-\dfrac{5}{18}=\dfrac{6-5}{18}=\dfrac{1}{18}\)
c) \(\dfrac{2}{5}-\dfrac{3}{20}=\dfrac{8}{20}-\dfrac{3}{20}=\dfrac{8-3}{20}=\dfrac{5}{20}=\dfrac{1}{4}\)
\(\dfrac{1}{3}+\dfrac{2}{9}=\dfrac{3}{3\times3}+\dfrac{2}{9}=\dfrac{3}{9}+\dfrac{2}{9}=\dfrac{5}{9}\)
\(\dfrac{1}{2}+\dfrac{3}{8}=\dfrac{4}{2\times4}+\dfrac{3}{8}=\dfrac{4}{8}+\dfrac{3}{8}=\dfrac{7}{8}\)
\(\dfrac{5}{12}+\dfrac{2}{3}=\dfrac{5}{12}+\dfrac{2\times4}{3\times4}=\dfrac{5}{12}+\dfrac{8}{12}=\dfrac{13}{12}\)
\(\dfrac{5}{16}+\dfrac{3}{8}=\dfrac{5}{16}+\dfrac{3\times2}{8\times2}=\dfrac{5}{16}+\dfrac{6}{16}=\dfrac{11}{16}\)
\(\dfrac{4}{15}+\dfrac{3}{5}=\dfrac{4}{15}+\dfrac{3\times3}{5\times3}=\dfrac{4}{15}+\dfrac{9}{15}=\dfrac{13}{15}\)
\(\dfrac{8}{63}+\dfrac{7}{10}=\dfrac{8\times10}{63\times10}+\dfrac{7\times63}{10\times63}=\dfrac{80}{630}+\dfrac{441}{630}=\dfrac{521}{630}\)
1/ \(\left(\dfrac{2021}{2020}+\dfrac{2020}{2021}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
=\(\left(\dfrac{2021}{2020}+\dfrac{2020}{2021}\right).0\)
=\(0\)
g: \(=\dfrac{-3}{4}-\dfrac{1}{4}+\dfrac{5}{7}+\dfrac{2}{7}+\dfrac{3}{5}=\dfrac{3}{5}\)
h: \(=\dfrac{7}{19}\left(\dfrac{8}{11}+\dfrac{3}{11}\right)-\dfrac{12}{19}=\dfrac{7}{19}-\dfrac{12}{19}=-\dfrac{5}{19}\)
i: \(=\dfrac{2013}{7}\left(19+\dfrac{5}{8}-26-\dfrac{5}{8}\right)=\dfrac{2013}{7}\cdot\left(-7\right)=-2013\)
a) Ta có: \(\left(\dfrac{617}{191}+\dfrac{29}{33}-\dfrac{115}{17}\right)\cdot\left(\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{20}\right)\)
\(=\left(\dfrac{617}{191}+\dfrac{29}{33}-\dfrac{115}{17}\right)\cdot\left(\dfrac{5}{20}-\dfrac{4}{20}-\dfrac{1}{20}\right)\)
\(=0\cdot\left(\dfrac{617}{191}+\dfrac{29}{33}-\dfrac{115}{17}\right)=0\)
b) Ta có: \(\dfrac{12}{5}\cdot\left(\dfrac{10}{3}-\dfrac{5}{12}\right)\)
\(=\dfrac{12}{5}\cdot\left(\dfrac{40}{12}-\dfrac{5}{12}\right)\)
\(=\dfrac{12}{5}\cdot\dfrac{35}{12}\)
=7
\(a,=\dfrac{5}{7}-\dfrac{33}{8}=-\dfrac{191}{56}\\ b,=\left(\dfrac{12}{17}+\dfrac{5}{17}\right)+\left(\dfrac{19}{7}+\dfrac{3}{7}\right)=1+3=4\\ c,=\left(0,125\cdot8\right)^{12}-\left(\dfrac{45}{15}\right)^3=1-3^3=-26\\ d,=\left(-\dfrac{1}{3}\right)\left(5\dfrac{2}{7}-2\dfrac{2}{7}\right)=-\dfrac{1}{3}\cdot3=-1\\ e,=\dfrac{3^4\cdot3^6}{3^9}=3\)
\(\dfrac{2}{5}+\dfrac{1}{5}=\dfrac{2+1}{5}=\dfrac{3}{5}\)
\(\dfrac{2}{3}+\dfrac{5}{3}=\dfrac{2+5}{3}=\dfrac{7}{3}\)
\(\dfrac{3}{8}+\dfrac{4}{8}=\dfrac{3+4}{8}=\dfrac{7}{8}\)
\(\dfrac{6}{9}+\dfrac{2}{9}=\dfrac{6+2}{9}=\dfrac{8}{9}\)
\(\dfrac{12}{18}+\dfrac{7}{18}=\dfrac{12+7}{18}=\dfrac{19}{18}\)
\(\dfrac{7}{4}+\dfrac{2}{4}=\dfrac{7+2}{4}=\dfrac{9}{4}\)
Các bạn không cần trả lời câu hỏi trên của mik vì mik đã hiểu rồi nha . Cho nên đừng trả lời ! OK
\(\dfrac{-2}{3}.\dfrac{-5}{8}\) = \(\dfrac{5}{12}\)