em lớp 6  nha

B= 1/2 + 1/6 + 1/12 +1/20 + 1/30 + 1/42 + 1/56 + 1/72

B= 1/1*2 + 1/2*3 + 1/3*4 + 1/4*5 + 1/5*6 + 1/6*7 + 1/7*8 + 1/8*9

B=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9

B=1+0-0-0-0-0-0-0-1/9

B=1-1/9

B=8/9

k em nha

phần A em chịu

còn câu dưới đó e

\(A=-\dfrac{1}{20}+\dfrac{-1}{30}+\dfrac{-1}{42}+\dfrac{-1}{56}+\dfrac{-1}{72}+\dfrac{-1}{90}\\ A=-\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\\ A=-\left(\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+\dfrac{8-7}{7.8}+\dfrac{9-8}{8.9}+\dfrac{10-9}{9.10}\right)\\ A=-\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ A=-\left(\dfrac{1}{4}-\dfrac{1}{10}\right)=-\dfrac{3}{20}.\)

Ta có: \(B=\dfrac{1}{6}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)

\(\Leftrightarrow B=\dfrac{1}{6}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)

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

\(\Leftrightarrow B=\dfrac{1}{6}+\dfrac{1}{4}-\dfrac{1}{10}=\dfrac{19}{60}\)

\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)

\(=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=\dfrac{1}{2}-\dfrac{1}{10}\)

\(=\dfrac{2}{5}\)

Mình nghĩ đề bài của bạn thiếu mất một phân số đó là \(\dfrac{1}{12}\) nhé. Nếu không có phân số đấy thì số nó hơi to một tí.

Sửa đề: A=-1/20+(-1/30)+(-1/42)+(-1/56)+(-1/72)+(-1/90)

=-(1/20+1/30+...+1/90)

=-(1/4-1/5+1/5-1/6+...+1/9-1/10)

=-1/4+1/10

=-5/20+2/20=-3/20

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

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

`=1/[4xx5]+1/[5xx6]+1/[6xx7]+...+1/[11xx12]`

`=1/4-1/5+1/5-1/6+1/6-1/7+...+1/11-1/12`

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

\(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\\ =\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}+\dfrac{1}{8\times9}+\dfrac{1}{9\times10}+\dfrac{1}{10\times11}+\dfrac{1}{11\times12}\\ =\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\\ =\dfrac{1}{4}-\dfrac{1}{12}\\ =\dfrac{3}{12}-\dfrac{1}{12}=\dfrac{2}{12}=\dfrac{1}{6}\)

D=\(-\dfrac{1}{4.5}\)+(\(-\dfrac{1}{5.6}\))+(\(-\dfrac{1}{6.7}\))+(\(-\dfrac{1}{7.8}\))+(\(-\dfrac{1}{8.9}\))+(\(-\dfrac{1}{9.10}\))

D=\(-\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\)

D=\(-\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

D=\(-\left(\dfrac{1}{4}-\dfrac{1}{10}\right)\)

D=\(-\dfrac{3}{20}\)

\(B=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)

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

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

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

\(B=-\left(\frac{1}{4}-\frac{1}{10}\right)=-\frac{3}{20}\)

=>B=\(-\left(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}\right)\)

=>B=\(-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+..+\frac{1}{9}-\frac{1}{10}\right)\)

=>B=\(-\left(\frac{1}{4}-\frac{1}{10}\right)\) =\(\frac{-3}{40}\) HẾTTTTTTTTTTTTT

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

\(b=\frac{-1}{4.5}+\frac{-1}{5.6}+\frac{-1}{6.7}+\frac{-1}{7.8}+\frac{-1}{8.9}+\frac{-1}{9.10}\)

\(b=\frac{-1}{4}-\frac{-1}{5}+\frac{-1}{5}-\frac{-1}{6}+\frac{-1}{6}+.......+\frac{-1}{9}+\frac{-1}{9}-\frac{-1}{10}\)

\(b=\frac{-1}{4}-\frac{-1}{10}\)

\(b=\frac{-3}{20}\)