\(M=\left(\dfrac{0,4-\dfrac{2}{9}+\dfrac{2}{11}}{1,4-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-0,25+\dfrac{1}{5}}{1\dfrac{1}{6}-0,875+0,7}\right):\dfrac{2012}{2013}\)

\(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{2}{6}-\dfrac{2}{8}+\dfrac{2}{10}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\right):\dfrac{2012}{2013}\)

\(M=\left(\dfrac{2\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}-\dfrac{2\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}{7\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}\right):\dfrac{2012}{2013}\)

\(M=\left(\dfrac{2}{7}-\dfrac{2}{7}\right).\dfrac{2013}{2012}\)

\(M=0.\dfrac{2013}{2012}\)

\(M=0\)

`a)` Xét tử số phân số M :

\(2012-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{2012}{2020}\\ =\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{2012}{2020}\right)\\ =\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{2020}\\ =24\left(\dfrac{1}{27}+\dfrac{1}{30}+\dfrac{1}{33}+...+\dfrac{1}{6060}\right)\)

Ta được : \(M=\dfrac{24\left(\dfrac{1}{27}+\dfrac{1}{30}+\dfrac{1}{33}+...+\dfrac{1}{6060}\right)}{\dfrac{1}{27}+\dfrac{1}{30}+\dfrac{1}{33}+...+\dfrac{1}{6060}}=24\)

`b)` Xét tử số phân số N :

\(\dfrac{1}{1.300}+\dfrac{1}{2.301}+\dfrac{1}{3.302}+...+\dfrac{1}{101.400}\\ =\dfrac{1}{299}.\left(\dfrac{299}{1.300}+\dfrac{299}{2.301}+\dfrac{299}{3.302}+...+\dfrac{299}{101.400}\right)\\ =\dfrac{1}{299}.\left(1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+\dfrac{1}{3}-\dfrac{1}{302}+...+\dfrac{1}{101}-\dfrac{1}{400}\right)\)

Xét mẫu số phân số N :

\(\dfrac{1}{1.102}+\dfrac{1}{2.103}+\dfrac{1}{3.104}+...+\dfrac{1}{299.400}\\ =\dfrac{1}{101}.\left(\dfrac{101}{1.102}+\dfrac{101}{2.103}+\dfrac{101}{3.104}+...+\dfrac{101}{299.400}\right)\\ =\dfrac{1}{101}.\left(1-\dfrac{1}{102}+\dfrac{1}{2}-\dfrac{1}{103}+\dfrac{1}{3}-\dfrac{1}{104}+...+\dfrac{1}{299}-\dfrac{1}{400}\right)\)

\(=\dfrac{1}{101}.\left(1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+\dfrac{1}{3}-\dfrac{1}{302}+...+\dfrac{1}{101}-\dfrac{1}{400}\right)\)

Ta được: \(N=\dfrac{\dfrac{1}{299}\left(1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+\dfrac{1}{3}-\dfrac{1}{302}+...+\dfrac{1}{101}-\dfrac{1}{400}\right)}{\dfrac{1}{101}\left(1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+\dfrac{1}{3}-\dfrac{1}{302}+...+\dfrac{1}{101}-\dfrac{1}{400}\right)}\\ =\dfrac{\dfrac{1}{299}}{\dfrac{1}{101}}=\dfrac{101}{299}\)

thank kiu ak

giúp mình với mai thi rùi

a: =4/5+1/5+2/3+1/3=1+1=2

b: =17/12+7/12+29/7-8/7=3+2=5

c: =3/5+2/5+16/7-1/7-1/7

=1+2=3

d: =2/5+3/5+2/3+1/3+7/4+1/4

=1+1+2

=4

\(\dfrac{4}{5}+\dfrac{2}{3}+\dfrac{1}{5}+\dfrac{1}{3}\)

\(=\left(\dfrac{4}{5}+\dfrac{1}{5}\right)+\left(\dfrac{2}{3}+\dfrac{1}{3}\right)\)

\(=\dfrac{5}{5}+\dfrac{3}{3}\)

\(=1+1\)

\(=2\)

============

\(\dfrac{17}{12}+\dfrac{29}{7}-\dfrac{8}{7}+\dfrac{7}{12}\)

\(=\left(\dfrac{17}{12}+\dfrac{7}{12}\right)+\left(\dfrac{29}{7}-\dfrac{8}{7}\right)\)

\(=\dfrac{24}{12}+\dfrac{21}{7}\)

\(=2+3\)

\(=5\)

====================

\(\dfrac{9}{15}+\dfrac{16}{7}+\dfrac{2}{5}-\dfrac{1}{7}-\dfrac{2}{14}\)

\(=\dfrac{9}{15}+\dfrac{16}{7}+\dfrac{6}{15}-\dfrac{1}{7}-\dfrac{1}{7}\)

\(=\left(\dfrac{9}{15}+\dfrac{6}{15}\right)+\left(\dfrac{16}{7}-\dfrac{1}{7}-\dfrac{1}{7}\right)\)

\(=\dfrac{15}{15}+\dfrac{14}{7}\)

\(=1+2\)

\(=3\)

===============

\(\dfrac{2}{5}+\dfrac{6}{9}+\dfrac{7}{4}+\dfrac{3}{5}+\dfrac{1}{3}+\dfrac{1}{4}\)

\(=\dfrac{2}{5}+\dfrac{2}{3}+\dfrac{7}{4}+\dfrac{3}{5}+\dfrac{1}{3}+\dfrac{1}{4}\)

\(=\left(\dfrac{2}{5}+\dfrac{3}{5}\right)+\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(\dfrac{7}{4}+\dfrac{1}{4}\right)\)

\(=\dfrac{5}{5}+\dfrac{3}{3}+\dfrac{8}{4}\)

\(=1+1+2\)

\(=4\)