1/30 + 1/42 +1/56 +1/72+1/90+1/110+1/132

= 1/5x6+1/6x7+1/7x8+1/8x9+1/9x10+1/10x11+1/11x12

=1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11+1/11-1/12

= 1/5 -1/12

=7/60

làm sai rồi lấy đâu ra dấu trừ phải là 1/5+ 1/12 chứ

bạn Đặng DuyPhương làm sai rồi

\(\Rightarrow\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}+...+\dfrac{1}{11\times12}=\dfrac{21}{x}\\ \Rightarrow\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{12}=\dfrac{21}{x}\\ \Rightarrow\dfrac{1}{5}-\dfrac{1}{12}=\dfrac{21}{x}\\ \Rightarrow\dfrac{21}{x}=\dfrac{7}{60}\Rightarrow x=\dfrac{21\cdot60}{7}=180\)

\(\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}=\dfrac{21}{x}\)

\(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}=\dfrac{21}{x}\)

\(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}=\dfrac{21}{x}\)

\(\dfrac{1}{5}-\dfrac{1}{12}=\dfrac{21}{x}\)

Còn lại bạn tự tính

\(B=\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}+\dfrac{1}{156}\)

\(B=\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}+\dfrac{1}{12\cdot13}\)

\(B=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}\)

\(B=\dfrac{1}{5}-\dfrac{1}{13}=\dfrac{8}{65}\)

B=\(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}+\dfrac{1}{12.13}\)

B=\(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-.......-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\)

B=\(\dfrac{1}{5}-\dfrac{1}{12}=\dfrac{7}{60}\)

(sở dĩ có như trên vì \(\dfrac{1}{x\left(x+1\right)}=\dfrac{1}{x}-\dfrac{1}{x+1}\) nên cứ áp dụng vào là ra  nhé)

Ta viết lại biểu thức A như sau:

\(A=-\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{11.12}\right)\)

\(A=-\left(\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+...+\dfrac{12-11}{11.12}\right)\)

\(A=-\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{12}\right)\)

\(A=-\left(\dfrac{1}{4}-\dfrac{1}{12}\right)\)

\(A=-\dfrac{1}{6}\)

B = 1/30 + 1/42 + 1/56 + 1/72 + 1/90 + 1/110 + 1/132 + 1/156

B = 1/5x6 + 1/6x7 + 1/7x8 + 1/8x9 + 1/9x10 + 1/10x11 + 1/11x12 + 1/12x13

B = 1/5 -1/6 + 1/6- 1/7 + 1/7  -1/ 8 + 1/8 -1/9 +1/9 -1/10 + 1/10  - 1/11 + 1/11-1/12 +1/12 -1/13

B = 1/5 - 1/13 

B = 8/65

\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)

\(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)

Ta có: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\) với mọi số tự nhiên n

\(\Rightarrow A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)

\(A=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)

Vậy A=7/60
 

A=\(\frac{1}{5.6}\)+\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)+\(\frac{1}{8.9}\)+\(\frac{1}{9.10}\)+\(\frac{1}{10.11}\)+\(\frac{1}{11.12}\)

  =1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11+1/11-1/12

  =1/5-1/12

  =7/60

Dấu chấm là dấu nhân nhé bạn

A=1/30+1/42+1/56+1/72+1/90+1/110+1/132

A=1/5*6+1/6*7+1/7*8+1/8*9+1/9*10+1/10*11+1/11*12

A=1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11+1/11-1/12

A=1/5-1/12

A=7/60

A=\(\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\)

A=\(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}\)

A=\(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\)

A=\(\dfrac{1}{5}-\dfrac{1}{12}\)

A=\(\dfrac{12}{60}+\dfrac{-5}{60}\)

A=\(\dfrac{7}{60}\)

\(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{11.12}\)

\(;A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)

\(;A=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)

=1/5.6+1/6.7+1/7.8+`1/8.9+1/9.10+1/10.11+1/11.12

=1/5-1/12

=7/60

\(\frac{1}{5x6}+\frac{1}{6x7}+\frac{1}{7x8}+\frac{1}{8x9}+\frac{1}{9x10}+\frac{1}{10x11}+\frac{1}{11x12}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)

  =1/5-1/12=7/60

B= 1/30 + 1/42 + 1/56 + 1/72 + 1/90 + 1/110 + 1/132= 7/60

\(\dfrac{1}{20}=\dfrac{1}{4x5}=\dfrac{1}{4}-\dfrac{1}{5}\)

Tương tự các phân số khác

S= \(\dfrac{1}{4}-\dfrac{1}{12}=\dfrac{1}{6}\)

\(\dfrac{1}{20}+\dfrac{1}{30}\)+ \(\dfrac{1}{42}\)+\(\dfrac{1}{56}\)+\(\dfrac{1}{72}\)+\(\dfrac{1}{90}\)+\(\dfrac{1}{110}\)+\(\dfrac{1}{132}\)

= \(\dfrac{1}{4\times5}\)+\(\dfrac{1}{5\times6}\)+\(\dfrac{1}{6\times7}\)+\(\dfrac{1}{7\times8}\)+\(\dfrac{1}{8\times9}\)+\(\dfrac{1}{9\times10}\)+\(\dfrac{1}{10\times11}\)+\(\dfrac{1}{11\times12}\)

= \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)+\(\dfrac{1}{9}\)-\(\dfrac{1}{10}\)+\(\dfrac{1}{10}\)-\(\dfrac{1}{11}\)+\(\dfrac{1}{11}\)-\(\dfrac{1}{12}\)

= \(\dfrac{1}{4}\) - \(\dfrac{1}{12}\)

= \(\dfrac{3}{12}\) - \(\dfrac{1}{12}\)

= \(\dfrac{2}{12}\)

=\(\dfrac{1}{6}\)

=1/4x5+1/5x6+1/6x7+1/7x8+1/8x9+1/9x10+1/10x11+1/11x12

=1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11+1/11-1/12

=1/4-1/12=3/12-1/12=2/12=1/6