\(\frac{1}{2}\)+   \(\frac{1}{6}\)+   \(\frac{1}{12}\)+  \(\frac{1}{20}\)+   \(\frac{1}{30}\)+   \(\frac{1}{42}\)+    \(\frac{1}{56}\)

=   \(\frac{1}{1.2}\)+   \(\frac{1}{2.3}\)+    \(\frac{1}{3.4}\)+   \(\frac{1}{4.5}\)+   ......   +   \(\frac{1}{7.8}\)

=   \(1\)\(-\)\(\frac{1}{8}\)

=    \(\frac{7}{8}\)

thiếu bước :v

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\)

\(=1-\frac{1}{8}\)

\(=\frac{7}{8}\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

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

\(=1-\frac{1}{8}\)

\(=\frac{7}{8}\)

A = \(\frac{-79}{90}\)

B = \(\frac{8}{9}\)

cách giải sao chỉ mình với

\(\frac{6}{7}\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)

=\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}\)

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

=\(1-\frac{1}{7}\)=\(\frac{6}{7}\)

\(\frac{6}{7}\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

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

\(=\frac{9}{10}\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)

= \(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{9\times10}\)

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

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

1/2 + 1/6 + 1/12 + 1/20 + .... + 1/90

= 1/1x2 + 1/2x3 + 1/3x4 + 1/4x5 + ..... + 1/9x10

= 2-1/1x2 + 3-2/2x3 + 4-3/3x4 + ...... + 10-9/9x10

= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ...... + 1/9 - 1/10

= 1- 1/10

= 9/10

k đúng cho mk nha mn ^.^

\(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-...-\frac{1}{2}\)

= \(\frac{8}{9}-\left(\frac{1}{72}+\frac{1}{56}+...+\frac{1}{2}\right)\)

= \(\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{72}\right)\)

= \(\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\right)\)

= \(\frac{8}{9}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\right)\)

= \(\frac{8}{9}-\left(1-\frac{1}{9}\right)\)

= \(\frac{8}{9}-\frac{8}{9}\)

= \(0\)

Chúc bạn học tốt !!!

Trả lời:

\(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

\(=\frac{8}{9}-\left(\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)

\(=\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)\)

\(=\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)

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

\(=\frac{8}{9}-\left(\frac{1}{1}-\frac{1}{9}\right)\)

\(=\frac{8}{9}-\frac{8}{9}\)

\(=0\)

1/2+1/6+1/12+...+1/110

=1/1.2+1/2.3+1/3.4+...+1/10.11

=1-1/2+1/2-1/3+1/3-1/4+...+1/10-1/11

=1-1/11=10/11

1/2

từ đầu bài ta có

1 phần 6 cộng một phần12 cộng..... cộng một phần 42

=1phần 2.3 +1 phần 3.4 +....+1 phần6.7

=1 phần2 -(1 phần 3 +1 phần 3)...........-(1 phần 6 +1 phần 6)-1 phần 7

1 phần 2 -1 phần 7

=5 phần 14 

mongcasc bạn giúp mk mai mk ktra chương, tất nhiên mk sẽ k cho ai nhanh mà đúng

Ta có:

\(\frac{1}{n\left(n+1\right)}=\frac{n+1-n}{n\left(n+1\right)}=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

\(\Rightarrow A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)

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

\(A=\frac{1}{2}-\frac{1}{7}\)

\(A=\frac{5}{14}\)

\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{30}+...+\frac{1}{72}+\frac{1}{81}\)

\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{8\times9}+\frac{1}{81}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{81}\)

\(A=1-\frac{1}{9}+\frac{1}{81}=\frac{73}{81}\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{81}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}+\frac{1}{81}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{81}\)

\(=1-\frac{1}{9}+\frac{1}{81}\)

\(=\frac{8}{9}+\frac{1}{81}\)

\(=\frac{73}{81}\)

\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{8\cdot9}+\frac{1}{81}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{81}\)
\(=1-\frac{1}{9}+\frac{1}{81}\)
\(=\frac{8}{9}+\frac{1}{81}\)
\(=\frac{73}{81}\)

\(S=\frac{3}{4}-0,25-\left[\frac{7}{3}+\left(\frac{-9}{2}\right)\right]-\frac{5}{6}\)

\(S=\frac{3}{4}-\frac{1}{4}-\left[\frac{14}{6}+\left(\frac{-27}{6}\right)\right]-\frac{5}{6}\)

\(S=\frac{1}{2}-\left(\frac{-13}{6}\right)-\frac{5}{6}\)

\(S=\frac{3}{6}-\left(\frac{-13}{6}\right)-\frac{5}{6}\)

\(S=\frac{11}{6}\)

\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)

\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(B=1-\frac{1}{100}\)

\(B=\frac{99}{100}\)

\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\) 

  \(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

  \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

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

    \(=1-\frac{1}{100}=\frac{99}{100}\)

\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\)

\(\Rightarrow B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

\(\Rightarrow B=1-\frac{1}{100}\)

\(\Rightarrow B=\frac{100}{100}-\frac{1}{100}\)

\(\Rightarrow B=\frac{99}{100}\)

Vậy \(B=\frac{99}{100}\)