Ta có:

\(A=\frac{3}{1\cdot5}+\frac{3}{5\cdot10}+...+\frac{3}{100\cdot105}\)

\(=\frac{3}{5}\cdot\left(\frac{5}{1\cdot5}+\frac{5}{5\cdot10}+...+\frac{5}{100\cdot105}\right)\)

\(=\frac{3}{5}\cdot\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{105}\right)\)

\(=\frac{3}{5}\left(1-\frac{1}{105}\right)=\frac{3}{5}\cdot\frac{104}{105}=\frac{312}{525}\)

\(=\dfrac{4}{7}\)

\(B=\dfrac{5}{1.2}+\dfrac{13}{2.3}+\dfrac{25}{3.4}+\dfrac{41}{4.5}+...+\dfrac{181}{9.10}\)

\(=\left(\dfrac{1}{1.2}+\dfrac{4}{1.2}\right)+\left(\dfrac{1}{2.3}+\dfrac{12}{2.3}\right)+\left(\dfrac{1}{3.4}+\dfrac{24}{3.4}\right)+...+\left(\dfrac{1}{9.10}+\dfrac{180}{9.10}\right)\)

\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\right)+\left(\dfrac{4}{1.2}+\dfrac{12}{2.3}+...+\dfrac{180}{9.10}\right)\)

\(=\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)+\left(2+2+...+2\right)\)

\(=1-\dfrac{1}{10}+\left(2.9\right)\)

\(=1-\dfrac{1}{10}+18\)

\(=\dfrac{9}{10}+18\)

\(=18\dfrac{9}{10}\)

a) \(\dfrac{30\times25\times7\times8}{75\times8\times12\times14}=\dfrac{3\times2\times5\times25\times7\times8}{25\times3\times8\times3\times4\times2\times7}=\dfrac{5}{3\times4}=\dfrac{5}{12}\)

b) \(\dfrac{8\times3\times4}{16\times3}=\dfrac{8\times3\times2\times2}{8\times2\times3}=2\)

c) \(\dfrac{4\times5\times6}{3\times10\times8}=\dfrac{4\times5\times3\times2}{3\times5\times2\times4\times2}=\dfrac{1}{2}\)

giúp mik với ạ, mình sẽ tick ạ! Thanks.

\(B=\dfrac{40404}{70707}+\dfrac{244\times395-151}{244+395\times243}+\dfrac{1\times3\times5+2\times6\times10+4\times12\times20+7\times21\times35}{1\times5\times7+2\times10\times14+4\times20\times28+7\times35\times49}\\ =\dfrac{4}{7}+\dfrac{243\times395+395-151}{244+395\times243}+\dfrac{1\times3\times5\left(1+2+4+7\right)}{1\times5\times7\left(1+2+4+7\right)}\\ =\dfrac{4}{7}+\dfrac{243\times395+244}{244+395\times243}+\dfrac{3}{7}\\ =\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+1\\ =1+1=2\)

a: \(=\left(\dfrac{1}{15}+\dfrac{14}{15}\right)+\left(\dfrac{9}{10}-2-\dfrac{11}{9}\right)+\dfrac{1}{157}\)

\(=1+\dfrac{1}{157}+\dfrac{81-180-110}{90}\)

\(=\dfrac{158}{157}+\dfrac{-209}{90}\simeq-1.315\)

b: \(=\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{2}{6}\)

=1/3-1/3

=0

c: \(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2015\cdot2017}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)

=2016/2017

\(\dfrac{10}{11}:\left(\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+\dfrac{2}{9\times11}\right)\)

\(=\dfrac{10}{11}:\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)\)

\(=\dfrac{10}{11}:\left(\dfrac{1}{3}-\dfrac{1}{11}\right)\)

\(=\dfrac{10}{11}:\dfrac{8}{33}\)

\(=\dfrac{10}{11}\times\dfrac{33}{8}\)

\(=5\times\dfrac{3}{4}\)

\(=\dfrac{15}{4}\)

A =\(\dfrac{4^2}{3\times5}\) \(\times\)\(\dfrac{5^2}{4\times6}\) \(\times\) \(\dfrac{6^2}{5\times7}\) \(\times\) \(\dfrac{7^2}{6\times8}\)

A = \(\dfrac{4\times4\times5^2\times6^2\times7\times7}{3\times4\times5^2\times6^2\times7\times8}\)

A =  \(\dfrac{4}{3}\) \(\times\) \(\dfrac{7}{8}\)

A = \(\dfrac{7}{6}\) 

S = 1.3 + 3.5 + 5.7 + ...+ 99.101

=>6S = 1.3.6 + 3.5.6 + 5.7.6 + ...+ 99.101.6

6S = 1.3.(5+1) + 3.5.(7-1) + 5.7.(9-3) + ...+ 99.101.(103-97)

6S = 1.3.5 + 1.3 + 3.5.7 - 1.3.5 + 5.7.9 - 3.5.7 + ...+ 99.101.103 - 97.99.101

6S = 1.3 + 99.101.103

S = 171 650

S = 1.3 + 3.5 + 5.7 + ...+ 99.101

=>6S = 1.3.6 + 3.5.6 + 5.7.6 + ...+ 99.101.6

6S = 1.3.(5+1) + 3.5.(7-1) + 5.7.(9-3) + ...+ 99.101.(103-97)

6S = 1.3.5 + 1.3 + 3.5.7 - 1.3.5 + 5.7.9 - 3.5.7 + ...+ 99.101.103 - 97.99.101

6S = 1.3 + 99.101.103

S = 171 650