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) \(B=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}\)

      \(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}-\frac{1}{8}+\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)

       \(=\frac{1}{2}-\frac{1}{14}=\frac{3}{7}\)

b) Ta có : A = \(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{99.100}\)

                  \(=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

                   \(=3.\left(1-\frac{1}{100}\right)\)

                     \(=3.\frac{99}{100}=\frac{297}{100}\)

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

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

d)\(\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{97.99}\) 

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{99}\)

\(=\frac{1}{3}-\frac{1}{99}\)

\(=\frac{32}{99}\)

a)   11/4x(-0,4)x-1,6x11/4

     = -1,1 x-1,6x11/4

    = 1,76x11/4

    = 4,48

b)    6/19x-7/11+6/19x-4/19x-4/11+13/19

     = -42/209+96/3971+13/19

     = -702/3971+13/19

     = 2015/3971

c)     2/5x3/7+-3/7x3/5+3/7

      = 3/7x(2/5+3/5+1)

      = 3/7x(1+1)

      = 3/7x2

      = 6/7

d)      2/3x5+2x5/7+2/7x9+....+2/97x99

      = 1/3-1/5+1/5-1/7+1/7-1/9+......+1/97-1/99

      = 1/3-1/99

      = 32/99

d)=1/3-1/5+1/5-1/7+....+1/97-1/99

=1/3-1/99

=32/99

giúp mình bài này với

bạn nam đọc 1 quyển sách trong 3 ngày . ngày 1 đọc 1 phần 3 số trang . ngày thứ 2 đọc 2 phần 5 số trang còn lại .ngày 3 đọc 36 trang còn lại. hỏi quyển sách có bao nhiêu trang

mình giúp bạn thì bạn giúp mình nhé

 `A) 1/3 + 1/5 - 1/4 = 20/60 + 12/60 - 15/60 = 17/60`

` B) 7/8 - ( 1/4 + 2/5 ) = 7/8 - 2/4 -2/5 = 5/8-2/8 = 9/40`

` C) 5/11 - ( 3/5 - 6/11 ) = 5/11 - 3/55 = 25/55 - 3/55 = 22/55=2/5 ` 

` D) 5/7 - ( 19/35 - 6/15 ) = 5/7 - 1/7 = 4/7 `

`1/3 + 1/5 -1/4`

`= 5/15 + 3/15 -1/4`

`= 8/15 -1/4`

`=  17/60`

__

`7/8 -(1/4 + 2/5)`

`= 7/8-1/4 - 2/5`

`= 7/8 - 2/8 -2/5`

`= 5/8 -2/5`

`= 9/40`

__

`5/11-(3/5 -6/11)`

`= 5/11 -3/5 +6/11`

`= (5/11 + 6/11)-3/5`

`= 1 -3/5`

`= 5/5-3/5`

`=2/5`

__

`5/7 - ( 19/35 - 6/15)`

`= 5/7 - 19/35 + 6/15`

`= 25/35 -19/35 + 6/15`

`=6/35 +6/15`

`=4/7`

b: \(B=\dfrac{5}{2}-\dfrac{7}{2}+\dfrac{3}{8}+\dfrac{6}{8}+\dfrac{-6}{11}-\dfrac{5}{11}=-2-1+\dfrac{9}{8}=\dfrac{9}{8}-3=-\dfrac{15}{8}\)

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

d: \(D=\dfrac{4}{19}\left(\dfrac{-5}{6}-\dfrac{7}{12}\right)-\dfrac{40}{57}\)

\(=\dfrac{4}{19}\cdot\dfrac{-17}{12}-\dfrac{40}{57}=-1\)

e: \(E=\dfrac{1}{3}\left(\dfrac{4}{5}-\dfrac{9}{5}\right)+\dfrac{2}{3}=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)