Đặt \(A=\frac{1}{30}+\frac{1}{42}+...+\frac{1}{90}+\frac{1}{110}\)

                                         \(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)

                                      \(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)

                                    \(A=\frac{1}{5}-\frac{1}{11}=\frac{11-5}{55}=\frac{6}{55}\)

                                         Ủng hộ mk nha!!!

T = 1/30 + 1/42 + 1/56 + 1/72 + 1/90 + 1/110
T = 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.10 + 1/10.11
T = 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10 + 1/10 - 1/11
T = 1/5 - 1/11 (Cứ hai thằng cạnh nhau cộng lại bằng 0, chỉ còn thằng đầu và thằng cuối) 
T = (11-5)/55
T = 14/55
 

1/30 + 1/42 + 1/ 56 + 1/72 + 1/90 + 1/110

= 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.10 + 1/10.11

= 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10 + 1/10 - 1/11

=1/5 - 1/11

= 11/55 - 5/55

= (11 - 5)/55

= 6/55

\(\frac{-1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

= \(-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)

=\(-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

=\(-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)

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

\(\frac{-1}{90}-\frac{-1}{72}-\frac{-1}{56}-\frac{-1}{42}-\frac{-1}{30}-\frac{-1}{20}-\frac{-1}{12}-\frac{-1}{6}-\frac{-1}{2}\)

\(=\frac{-1}{10.9}-\frac{-1}{9.8}-\frac{-1}{8.7}-\frac{-1}{7.6}-\frac{-1}{6.5}-\frac{-1}{5.4}-\frac{-1}{4.3}-\frac{-1}{3.2}-\frac{-1}{2.1}\)

\(=\frac{79}{90}\)

1/100 - 1/90 - 1/72 - 1/56 - 1/42 - 1/30 - 1/20 - 1/12 - 1/6 - 1/2 = -0,833

  -1/20 - 1/30 - 1/42 - 1/56 - 1/72 - 1/90 
= -1/20( 1 + 2/3 + 10/21 + 5/14 + 5/18 + 2/9 ) 
= -1/20. [ (1 + 2/3) + (10/21 + 5/14) + (5/18 + 2/9) 
= -1/20.( 5/3 + 5/6 + 1/2 ) 
= -1/20.( 5/2 + 1/2 ) 
= -1/20.3 
= -3/20 

\(C=\frac{8}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

\(=\frac{8}{90}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)\)

\(=\frac{8}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)

\(=\frac{4}{45}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)

\(=\frac{4}{45}-\left(1-\frac{1}{9}\right)=\frac{4}{45}-\frac{8}{9}=\frac{4}{45}-\frac{40}{45}=\frac{-36}{45}=\frac{-4}{5}\)

Ta có: -1/90 - 1/72 - 1/56 - 1/42 - 1/30 - 1/20 - 1/12 - 1/6 - 1/2

= (-1).(1/90 + 1/72 + 1/56 + 1/42 + 1/30 + 1/20 + 1/12 + 1/6 + 1/2)

= (-1).(1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90)

= (-1).(1/1 - 1/2 + 1/2 - 1/3 + ... + 1/9 - 1/10)

= (-1).(1 - 1/10)

= (-1).(9/10)

= -9/10

a)

1/12 + 1/6 + 1/2 = (1+2+6)/12 = 9/12 = 3/4 
1/30 + 1/20 = (2+3)/60 = 5/60 = 1/12 
1/56 + 1/42 = 1/7(1/8+1/6) = 1/7 .(3+4)/24 = 1/24 
8/9- 1/72 = (8.8 - 1)/72 = 63/72 = 7/8 
1/12 + 1/24 = (2+1)/24 = 1/8 
7/8 - 1/8 = 6/8 = 3/4 
3/4 - 3/4 = 0 
 

b)

\(0,5+\frac{1}{3}+0,4+\frac{5}{7}+\frac{1}{6}-\frac{4}{35}\)

\(=\left(\frac{1}{2}+\frac{1}{3}+\frac{2}{3}+\frac{1}{6}\right)+\left(\frac{5}{7}-\frac{4}{35}\right)\)

\(=\frac{15+10+12+5}{30}+\frac{25-4}{35}\)

\(=\frac{7}{5}+\frac{3}{5}\)

\(=2\)

duy cute ơi,câu b đáng lẽ là 2/5 chứ

\(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}=\dfrac{8}{9}-\left(\dfrac{1}{8.9}+\dfrac{1}{7.8}+\dfrac{1}{6.7}+\dfrac{1}{5.6}+\dfrac{1}{4.5}+\dfrac{1}{3.4}+\dfrac{1}{2.3}+\dfrac{1}{1.2}\right)=\dfrac{8}{9}-\left(\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{2}-\dfrac{1}{3}+1-\dfrac{1}{2}\right)=\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)=\dfrac{8}{9}-\dfrac{8}{9}=0\)