xin lỗi nha

\(=\frac{2}{145}\left(\frac{1}{9}+1\right)-\frac{13}{3}.\frac{1}{145}\)

\(=\frac{2}{145}.\frac{10}{9}-\frac{13}{3}.\frac{1}{145}\)

\(=\frac{1}{145}+\frac{1}{145}.\frac{10}{9}-\frac{13}{3}.\frac{1}{145}\)

\(=\frac{1}{145}\left(\frac{10}{9}-\frac{13}{3}+1\right)\)

\(=\frac{1}{145}.-\frac{20}{9}\)

\(=-\frac{4}{261}\)

Mk bị nhầm chỗ 2. 145 thành 2/145 nhé. thank kiuf.

\(\frac{1}{9}.\frac{2}{145}-\frac{13}{3}.\frac{2}{245}+\frac{2}{145}\)

\(=\frac{2}{145}\left(\frac{1}{9}-\frac{13}{3}+1\right)\)

\(=\frac{2}{145}.\left(-\frac{29}{9}\right)\)

\(=-\frac{2}{45}\)

a) \(\frac{3}{2}.\frac{7}{2}+\frac{5}{3}.\frac{1}{2}=\frac{21}{2}+\frac{5}{6}=\frac{63}{6}+\frac{5}{6}=\frac{68}{6}=\frac{34}{3}\)

b) \(\frac{1}{9}.\frac{2}{145}-\frac{13}{3}.\frac{2}{145}+\frac{2}{145}=\frac{2}{145}\left(\frac{1}{9}-\frac{39}{9}+\frac{9}{9}\right)=\frac{2}{145}\left(-29\right)=-0.4\)

c) \(\left(-\frac{91}{12}\right):\frac{15}{7}-\frac{1}{18}:\frac{15}{7}+\frac{20}{9}:\frac{15}{7}=\frac{7}{15}\left(-\frac{91}{12}-\frac{1}{18}+\frac{20}{9}\right)=\frac{7}{15}\left(-\frac{65}{12}\right)=-\frac{91}{36}\)

a,\(\frac{3}{2}\cdot\frac{7}{3}+\frac{4}{3}\cdot\frac{1}{2}\)=\(\frac{7}{2}+\frac{2}{3}\)=\(\frac{25}{6}\)

b,\(\frac{2}{145}\cdot\left(\frac{1}{9}-\frac{13}{3}+1\right)\)=\(\frac{2}{145}.-\frac{29}{9}=-\frac{2}{45}\)

c.\(\frac{15}{7}:\left(-\frac{31}{12}-\frac{1}{18}-\frac{20}{9}\right)\)=\(\frac{15}{7}:\left(-\frac{175}{36}\right)=-\frac{108}{245}\)

\(1\frac{1}{2}.2\frac{1}{3}+1\frac{1}{3}.\frac{1}{2}\)

\(=\frac{3}{2}.\frac{7}{3}+\frac{4}{3}.\frac{1}{2}\)

\(=\frac{7}{2}+\frac{2}{3}\)

\(=\frac{25}{6}\)

\(\frac{1}{9}.\frac{2}{145}-4\frac{1}{3}.\frac{2}{145}+\frac{2}{145}\)

\(=\frac{2}{1305}-\frac{13}{3}.\frac{2}{145}+\frac{2}{145}\)

\(=\frac{2}{1305}-\frac{26}{435}+\frac{2}{145}\)

\(=\frac{-2}{45}\)

\(\left(-2\frac{7}{12}\right):2\frac{1}{7}-\frac{1}{18}:2\frac{1}{7}+2\frac{2}{9}:2\frac{1}{7}\)

\(=\frac{-17}{12}:\frac{15}{7}-\frac{1}{18}:\frac{15}{7}+\frac{20}{9}:\frac{15}{7}\)

\(=\left(\frac{-17}{2}-\frac{1}{18}+\frac{20}{9}\right):\frac{15}{7}\)

\(=\frac{-19}{3}.\frac{7}{15}\)

\(=\frac{-133}{45}\)

học tốt

a,    1 1/2 . 2 1/3 + 1 1/3 . 1/2

       =77/2+11/6

       =121/3

\(\dfrac{1}{9}.\dfrac{2}{145}-\dfrac{13}{3}.\dfrac{2}{145}+\dfrac{2}{145}\)

\(=\dfrac{2}{145}.\left(\dfrac{1}{9}-\dfrac{13}{3}+1\right)\)

\(=\dfrac{2}{145}.\left(\dfrac{1}{9}-\dfrac{13}{3}+\dfrac{9}{9}\right)\)

\(=\dfrac{2}{145}.\left(-\dfrac{29}{9}\right)\)\(=-\dfrac{2}{45}\)

c) \(\left(11+12+13+...+19\right).\left(6.8-48\right)=\left(11+12+13+...+19\right).\left(48-48\right)\)

                                                                                         \(=\left(11+12+13+...+19\right).0\)

                                                                                         \(=0\)

Hok "tuốt" nha^^

b) \(\left(500.9-250.18\right).\left(1+2+3+...+9\right)=\left(250.2.9-250.18\right).\left(1+2+...+9\right)\)

                                                                                             \(=\left(250.18-250.18\right).\left(1+2+3+...+9\right)\)

                                                                                             \(=0.\left(1+2+3+...+9\right)\)

                                                                                             \(=0\)

Hok tốt nha^^ (lần 2)

a) \(\dfrac{-5}{9}.\dfrac{3}{11}+\dfrac{-13}{18}.\dfrac{3}{11}\)

\(=\dfrac{3}{11}.\left(\dfrac{-5}{9}+\dfrac{-13}{9}\right)\)

\(=\dfrac{3}{11}.\left(-2\right)\)

\(=\dfrac{-6}{11}\)

b) \(\dfrac{11}{2}.2\dfrac{1}{3}-1\dfrac{1}{5}.1\dfrac{1}{2}\)

\(=\dfrac{11}{3}.\dfrac{7}{3}-\dfrac{6}{5}.\dfrac{3}{2}\)

\(=\dfrac{77}{9}-\dfrac{9}{5}\)

\(=\dfrac{385}{45}-\dfrac{81}{45}\)

\(=\dfrac{304}{45}\)

c) \(1\dfrac{1}{9}.\dfrac{2}{145}-4\dfrac{1}{3}-\dfrac{2}{145}+\dfrac{2}{145}\)

\(=\dfrac{10}{9}.\dfrac{2}{145}-\dfrac{8}{3}\)

\(=\dfrac{4}{261}-\dfrac{8}{3}\)

\(=\dfrac{4}{261}-\dfrac{696}{261}\)

\(=-\dfrac{692}{261}\)

d) \(1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}+4-\dfrac{1}{4}-3-\dfrac{1}{3}-2-\dfrac{1}{2}-1\)

\(=\left(1-1\right)+\left(2-2\right)+\left(3-3\right)+4-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)-\left(\dfrac{2}{3}+\dfrac{1}{3}\right)-\left(\dfrac{3}{4}+\dfrac{1}{4}\right)\)

\(=0+0+0+4-1-1-1\)

\(=4-3\)

\(=1\)

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