\(A=\frac{1}{1\cdot2}+\frac{2}{2\cdot4}+\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+...+\frac{10}{46\cdot56}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{56}\)

\(A=1-\frac{1}{56}\)

\(A=\frac{55}{56}\)

\(B=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}\)

\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)

\(B=\frac{1}{3}-\frac{1}{27}\)

\(B=\frac{8}{27}\)

\(C=\frac{4}{3\cdot6}+\frac{4}{6\cdot9}+\frac{4}{9\cdot12}+...+\frac{4}{99\cdot102}\)

\(C=\frac{4}{3}\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{99\cdot102}\right)\)

\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{102}\right)\)

\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{102}\right)\)

\(C=\frac{4}{3}\cdot\frac{33}{102}\)

\(C=\frac{22}{51}\)

Các bạn giải giúp mình nha😐

A.5/7

B.-481/391

C.0

D.-59/9

E.1

\(\dfrac{7}{-25}+\dfrac{18}{25}+\dfrac{4}{23}+\dfrac{5}{7}+\dfrac{19}{23}\)

=\(\left(\dfrac{-7}{25}+\dfrac{18}{25}\right)+\left(\dfrac{4}{23}+\dfrac{19}{23}\right)+\dfrac{5}{7}\)

= \(\dfrac{11}{25} +1+\dfrac{5}{7}\)

= \(1\dfrac{11}{25}+\dfrac{5}{7}\)

= \(2\dfrac{27}{175}\)

b) \(-2+\dfrac{15}{19}+\dfrac{-15}{17}+\dfrac{15}{23}+\dfrac{4}{19}\)

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

nhiều thế

nhieu the co ngia la ko bit 

a; \(\dfrac{3}{11}\) + \(\dfrac{5}{-9}\) + \(\dfrac{4}{11}\) - \(\dfrac{4}{9}\) + \(\dfrac{3}{17}\) + \(\dfrac{15}{11}\)

= (\(\dfrac{3}{11}\) + \(\dfrac{4}{11}\) + \(\dfrac{15}{11}\)) - (\(\dfrac{5}{9}\) + \(\dfrac{4}{9}\)) + \(\dfrac{3}{17}\)

= 2 - 1 + \(\dfrac{3}{17}\)

= 1 + \(\dfrac{3}{17}\)

= \(\dfrac{20}{17}\) 

c; N = \(\dfrac{\dfrac{5}{7}-\dfrac{5}{9}-\dfrac{5}{11}}{\dfrac{15}{7}+\dfrac{15}{9}+\dfrac{15}{11}}\)

   Phải là - \(\dfrac{5}{7}\) chỗ tử số mới đúng em nhé!

a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)

b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)

c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)

a: \(=\left(15-6-\dfrac{13}{18}\right):\dfrac{298}{27}-\dfrac{17}{8}:\dfrac{51}{40}\)

\(=\dfrac{149}{18}\cdot\dfrac{27}{298}-\dfrac{5}{3}=\dfrac{3}{2}-\dfrac{5}{3}=\dfrac{9-10}{6}=\dfrac{-1}{6}\)

b: \(=\dfrac{-16}{5}\cdot\dfrac{-15}{64}+\dfrac{-22}{15}:\dfrac{11}{2}\)

\(=\dfrac{3}{4}-\dfrac{4}{15}=\dfrac{29}{60}\)

c: \(=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+5+\dfrac{7}{9}=\dfrac{-7}{9}+\dfrac{7}{9}+5=5\)

d: \(=\dfrac{1}{2}\cdot\dfrac{4}{3}\cdot10\cdot\dfrac{1}{5}\cdot\dfrac{3}{4}=1\)

e: \(=\dfrac{4}{25}+\dfrac{11}{2}\cdot\dfrac{5}{2}+\dfrac{-23}{4}=\dfrac{204}{25}\)

\(1)\)\(-\dfrac{10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)

\(=\dfrac{10}{11}\left(-\dfrac{8}{9}+\dfrac{7}{18}\right)\)

\(=\dfrac{10}{11}\left(\dfrac{-16}{18}+\dfrac{7}{18}\right)\)

\(=\dfrac{10}{11}.\left(-\dfrac{1}{2}\right)=-\dfrac{5}{11}\)

\(2)\)\(\dfrac{12}{25}.\dfrac{23}{7}-\dfrac{12}{7}.\dfrac{13}{25}\)

\(=\dfrac{12}{7}.\dfrac{23}{25}-\dfrac{12}{7}.\dfrac{13}{25}\)

\(=\dfrac{12}{7}.\left(\dfrac{23}{25}-\dfrac{13}{25}\right)\)

\(=\dfrac{12}{7}.\dfrac{2}{5}=\dfrac{24}{35}\)

\(3)\)\(\dfrac{3}{7}.\dfrac{16}{15}-\dfrac{2}{15}.\dfrac{-3}{7}\)

\(=\dfrac{3}{7}.\dfrac{16}{15}-\dfrac{3}{7}.\dfrac{-2}{15}\)

\(=\dfrac{3}{7}.\left(\dfrac{16}{15}+\dfrac{2}{15}\right)\)

\(=\dfrac{3}{7}.\dfrac{18}{15}=\dfrac{18}{35}\)

\(4)\)\(-\dfrac{4}{13}.\dfrac{5}{17}+\dfrac{-12}{13}.\dfrac{4}{17}\)

\(=-\dfrac{4}{13}.\dfrac{5}{17}+\dfrac{-4}{13}.\dfrac{12}{17}\)

\(=-\dfrac{4}{13}.\left(\dfrac{5}{17}+\dfrac{12}{17}\right)\)

\(=-\dfrac{4}{13}.\dfrac{17}{17}=-\dfrac{4}{13}\)

`#040911`

`1)`

`-10/11 * 8/9 + 7/18 . 10/11`

`= 10/11 * (-8/9 + 7/18)`

`= 10/11 * (-1/2)`

`= -5/11`

`2)`

`12/25 * 23/7 - 12/7 *13/25`

`= 12/7 * 23/25 - 12/7 * 13/25`

`= 12/7 * (23/25 - 13/25)`

`= 12/7 * 2/5`

`= 24/35`

`3)`

`3/7 * 16/15 - 2/15 * (-3)/7`

`= 3/7 * (16/15 + 2/15)`

`= 3/7 * 6/5`

`= 18/35`

`4)`

`-4/13 * 5/17 + (-12)/13 * 4/17`

`= -4/17 * 5/13 + (-12)/13 * 4/17`

`= 4/17 * (-5/13 - 12/13)`

`= 4/17 * (-17)/13`

`= -4/13`

Mai tui giải cho hem

Ta có:

\(A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\)

\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\)

\(A=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}\)

\(B=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\)

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

\(B=2.\left(\frac{1}{1}-\frac{1}{10}\right)=2.\frac{9}{10}\)

\(B=\frac{9}{5}\)

tui con nho maaaaaaaa

??????????????/