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\(A=\frac{2019}{2}+\frac{2019}{6}+\frac{2019}{12}+....+\frac{2019}{2018.2019}\)
\(=\frac{2019}{1}.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2018.2019}\right)\)
\(=\frac{2019}{1}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\right)\)
\(=\frac{2019}{1}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+....+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(=\frac{2019}{1}.\left(1-\frac{1}{2019}\right)\)
\(=\frac{2019}{1}.\frac{2018}{2019}\)
\(=2018\)
\(A=\frac{2019}{2}+\frac{2019}{6}+\frac{2019}{12}+\frac{2019}{20}+\frac{2019}{30}+\frac{2019}{2018.2019}\)
\(A=\frac{2019}{1.2}+\frac{2019}{2.3}+\frac{2019}{3.4}+\frac{2019}{4.5}+\frac{2019}{5.6}+...+\frac{2019}{2018.2019}\)
\(A=2019.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\right)\)
\(A=2019.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(A=2019.\left(1-\frac{1}{2019}\right)\)\(=2019.\frac{2018}{2019}=2018\)
Vậy A = 2018
-Dấu " . " là dấu nhân.

1/6 + 1/12 + 1/20 + 1/30 + 1/42 + ... + 1/90 + 1/110 = 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + ... + 1/9.10 + 1/10.11 = 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + ... + 1/9 - 1/10 + 1/10 - 1/11 = 1/2 - 1/11 = 9/22
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+\frac{1}{110}\)
=\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
=\(\frac{1}{2}-\frac{1}{11}\)
=\(\frac{9}{22}\)

\(\frac{43}{20}+\frac{17}{6}:\left(\frac{5}{8}+\frac{7}{30}\right)\)
\(=\frac{43}{20}+\frac{17}{6}\cdot\frac{120}{103}\)
\(=\frac{43}{20}+\frac{340}{103}\)
\(=\frac{11229}{2060}\)


\(
\frac{48.700-24.45.20}{45-40+35-30+25-20+15-10+5}\)
\(=\frac{48.700-24.2.10.45}{5+5+5+5+5}\)
\(=\frac{48.700-48.450}{5.5}\)
\(=\frac{48.\left(700-450\right)}{25}\)
\(=\frac{48.250}{25}\)
\(=480\)

a) \(\left(\frac{4}{3}-\frac{4}{6}\right)+\left(\frac{4}{6}-\frac{4}{9}\right)+\left(\frac{4}{9}-\frac{4}{10}\right)+\left(\frac{4}{12}-\frac{4}{15}\right)\)
\(=\frac{4}{15}-\frac{4}{3}=\frac{-16}{15}\)
C) bạn chỉ ần bỏ các số giống nhau thôi nhé
= 1
b)

a) \(\frac{37}{25}+\frac{11}{21}-\frac{12}{25}+\frac{10}{21}\)
= \(\left(\frac{37}{25}-\frac{12}{25}\right)+\left(\frac{11}{21}+\frac{10}{21}\right)\)
= 1 + 1
= 2
\(\frac{37}{25}+\frac{11}{21}-\frac{12}{25}+\frac{10}{21}\)
=\(\left(\frac{37}{25}-\frac{12}{25}\right)+\left(\frac{11}{21}+\frac{11}{21}\right)\)
= 1 + 1 = 2 .
= 1.2 + 1 + 3 / 5.2
= 5.2 / 5.2
= 1