\(\dfrac{7}{4}.\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{333333}{303030}+\dfrac{33333333}{42424242}\right)\)

\(=\dfrac{7}{4}.\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)

\(=\dfrac{7}{4}.33\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right)\)

\(=\dfrac{7}{4}.33\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}\right)\)

\(=\dfrac{7}{4}.33\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)

\(=\dfrac{7}{4}.33\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)

\(=\dfrac{7}{4}.33\left(\dfrac{7}{21}-\dfrac{3}{21}\right)\)

\(=\dfrac{7}{4}.33.\dfrac{4}{21}\)

\(=\dfrac{231}{4}.\dfrac{4}{21}\)

\(=\dfrac{231}{21}=11\)

\(\frac{7}{4}\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)

=\(\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)

=\(\frac{7}{4}.33\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)

=\(\frac{231}{4}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)

=\(\frac{231}{4}\left(\frac{1}{3}-\frac{1}{7}\right)\)

=\(\frac{231}{4}.\frac{4}{21}\)

= 11

\(\dfrac{-7}{4}.\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{333333}{303030}+\dfrac{33333333}{42424242}\right)\)

\(=\dfrac{-7}{4}.\left(\dfrac{33}{12}.\dfrac{33.101}{20.101}.\dfrac{33.10101}{30.10101}.\dfrac{33.1010101}{42.1010101}\right)\)

\(=\dfrac{-7}{4}.\left(\dfrac{33}{12}.\dfrac{33}{20}.\dfrac{33}{30}.\dfrac{33}{42}\right)\)

\(=\dfrac{-7}{4}.33.\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right)\)

\(=\dfrac{-7}{4}.33.\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)

\(=\dfrac{-7}{4}.33.\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)

\(=\dfrac{-7}{4}.33.\dfrac{4}{21}\)

\(=33.\dfrac{-1}{3}\)

\(=11\)

xl nhé kết quả là -11 mk viết thiếu dấu -

\(P=-\frac{7}{4}.\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)

\(P=-\frac{7}{4}.\left(\frac{33}{12}+\frac{33.101}{20.101}+\frac{33x10101}{30x10101}+\frac{33x1010101}{42x1010101}\right)\)

\(P=-\frac{7}{4}.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)

\(P=-\frac{7}{4}.\left(\frac{11}{4}+\frac{33}{20}+\frac{11}{10}+\frac{11}{14}\right)\)

\(P=-\frac{77}{8}.\left(\frac{1}{6}+\frac{3}{10}+\frac{1}{15}+\frac{1}{21}\right)=-\frac{77}{8}.\frac{35+63+14+10}{210}=-\frac{11x122}{8x30}\)

\(P=-\frac{671}{120}\)

P = -7/4 x (33/12 + 3333/2020 + 333333/303030 + 33333333/42424242)
   = -7/4 x (33/12 + 33/20 + 33/30 + 33/42)
   = -7/4 x [33 x (1/12 + 1/20 + 1/30 + 1/42)]
   = -7/4 x [33 x (35/420 + 21/420 + 14/420 + 10/420)]
   = -7/4 x (33 x 80/420)
   = -7/4 x  33 x 4/21
   = -7/4 x 4/21 x 33 (= -7x4 / 4x21   x33)
   = -7/21 x 33 (= -7x33 / 21 = -1x7x3x11 / 3x7)
   = -11/1
   = -11
Đáp số P = -11

Các số gạch đi là do rút gọn phân số nhé!!!

Kết quả :    Viết lại biểu thức đã cho 

           =>         -7/4x . ( 33/12 + 33/20 + 33/30 + 33/42  ) = 22 

                    -7/4x . 33 . ( 1/12 + 1/20 + 1/30 + 1/42 )   = 22 

           -231/4x    . ( 1/3 . 4 + 1/ 4. 5 + 1/5 . 6 + 1/ 6. 7 ) = 22

-231/4x    . ( 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 ) = 22 

                                                -231/4x    . ( 1/3 - 1/7 ) = 22  

                                                -231/4x     .  4/21         = 22 

                                                                -11x            = 22 

                                                                    x             = 22 : -11

                                                                    x             = -2 

                                                      Vậy        x             = -2 

\(b,C=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\\ =\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+...+\dfrac{1}{30.33}\\ =\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+...+\dfrac{1}{30}-\dfrac{1}{33}\\ =\dfrac{1}{3}-\dfrac{1}{33}\\ =\dfrac{11}{33}-\dfrac{1}{33}=\dfrac{10}{33}\)

a.F=\(\dfrac{4}{2.4}\)+\(\dfrac{4}{4.6}\)+\(\dfrac{4}{6.8}\)+...+\(\dfrac{4}{2008.2010}\)

F=\(\dfrac{2.2}{2.4}\)+\(\dfrac{2.2}{4.6}\)+\(\dfrac{2.2}{6.8}\)+...+\(\dfrac{2.2}{2008.2010}\)

F=2.(\(\dfrac{2}{2.4}\)+\(\dfrac{2}{4.6}\)+\(\dfrac{2}{6.8}\)+...+\(\dfrac{2}{2008.2010}\))

F=2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{8}\)+...+\(\dfrac{1}{2008}\)-\(\dfrac{1}{2010}\))

F=2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{2010}\))

F=\(\dfrac{1004}{1005}\)

b, C=\(\dfrac{1}{18}\)+\(\dfrac{1}{54}\)+....+\(\dfrac{1}{990}\)

\(\Rightarrow\)C=\(\dfrac{1}{3.6}\)+\(\dfrac{1}{6.9}\)+...+\(\dfrac{1}{30.33}\)

=>3C=3( \(\dfrac{1}{3.6}\)+\(\dfrac{1}{6.9}\)+...+\(\dfrac{1}{30.33}\))

=>3C=\(\dfrac{3}{3.6}\)+\(\dfrac{3}{6.9}\)+....+\(\dfrac{3}{30.33}\)

=> 3C=\(\dfrac{1}{3}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{9}\)+...+\(\dfrac{1}{30}\)-\(\dfrac{1}{33}\)

=> 3C= \(\dfrac{1}{3}\)-\(\dfrac{1}{33}\)

=>3C=\(\dfrac{10}{33}\)

=> C=\(\dfrac{10}{33}\):3

=> C=\(\dfrac{10}{99}\)

a, F=4/2.4+4/4/6+4/6.8+......+4/2008.2010

=> F= 4/2.(2/2.4+2/4.6+2/6.8+......+2/2008/2010

=> F= 4/2. ( 1/2-1/4+1/4-1/6+1/6-1/8+......+1/2008-1/2010

=> F=4/2.( 1/2-1/2010)

=> F= 4/2. 502/1005

=> F= \(\dfrac{1004}{1005}\)

1, \(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)...........\left(1-\dfrac{1}{n+1}\right)\)

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

\(=\dfrac{1}{2}.\dfrac{2}{3}..............\dfrac{n}{n+1}\)

\(=\dfrac{1.2.3........n}{2.3.......\left(n+1\right)}\)

\(=\dfrac{1}{n+1}\)

2, \(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...........+\dfrac{1}{99.100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+............+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}\)

\(=\dfrac{99}{100}\)

C=\(-66\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{11}\right)+124.\left(-37\right)+63.\left(-124\right)\)

=\(-66.\left(\dfrac{5}{66}\right)+124\left(-37-63\right)=-5+124.\left(-100\right)\)

=-12405

\(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)...\left(1-\dfrac{1}{n+1}\right)\)

=\(\dfrac{1}{2}\).\(\dfrac{2}{3}\).\(\dfrac{3}{4}\).....\(\dfrac{n}{n+1}\)=\(\dfrac{1}{n+1}\)

Vậy...