Đề thiếu?

\(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{9900}\)

\(=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)

\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{4}-\frac{1}{100}\)

\(=\frac{6}{25}\)

1/20 + 1/30 + 1/42 + 1/56 +...+ 1/990                                                                                                                                                                 = 1/4.5 +1/5.6 +1/6.7 + 1/7.8 +...+ 1/99.100                                                                                                                                               =1/4 - 1/5 + 1/5 -1/6 + 1/6 -1/7 + 1/7 - 1/8 + ...+1/99 -1/100                                                                                                                          =1/4-1/100 = 24/100=6/25

=1/4.5+1/5.6+1/6.8+1/7.8+....1/33.30

=1/4-1/5+1/5-1/6+1/6-1/8+1/7-1/8+...+1/30-1/33

=1/4-1/33

=29/132

\(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{990}\)

\(=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)

\(=\frac{1}{4}-\frac{1}{100}\)

\(=\frac{6}{25}\)

1-1/990=989/990

em la hoc sinh lop 4 

\(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+.....+\frac{1}{990}\)

=\(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+......+\frac{1}{99.100}\)

=\(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+......+\frac{1}{99}-\frac{1}{100}\)(Phương pháp khử )

=\(\frac{1}{4}-\frac{1}{100}=\frac{100}{400}-\frac{4}{400}=\frac{96}{400}\)=\(\frac{6}{25}\)

Trần Thùy Dung nó đã bảo \(990\ne99\cdot100\) rùi mà vẫn tách như v

=\(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

=\(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

=\(\frac{1}{4}-\frac{1}{100}\)

=\(\frac{24}{100}=\frac{6}{25}\)

Bạn ơi sai rồi \(990\ne99.100\)

\(B1\)

\(=\frac{1}{1}-\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{37}-\frac{1}{38}-\frac{1}{39}\)

\(=1-\frac{1}{39}\)

\(=\frac{38}{39}\)

\(B2\)

\(=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+.....+\frac{1}{99\cdot100}\)

\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+......+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{4}-\frac{1}{100}\)

\(=\frac{25}{100}-\frac{1}{100}\)

\(=\frac{24}{100}\)

\(=\frac{6}{25}\)

Bài 1 :

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)

\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\)

\(=\frac{1}{1.2}-\frac{1}{38.39}\)

\(=\frac{370}{741}\)

Tớ chỉ biết làm bài 1 thui

1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ... + 1/37.38.39

= 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 - 1/3.4 + 1/3.4 - 1/4.5 + 1/37.38 - 1/38.39

= 1/1.2 - 1/38.39

= 370/741

A = \(\frac{-79}{90}\)

B = \(\frac{8}{9}\)

cách giải sao chỉ mình với

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

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

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

\(=\frac{1}{3}-\frac{1}{8}=\frac{5}{24}\)

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

\(=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)

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

\(=\frac{1}{3}-\frac{1}{8}\)

\(=\frac{8-3}{24}=\frac{5}{24}\)

1/12+1/20+1/30+1/42+1/56=5/24

t.i.c.k nha

\(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{9900}\)

\(A=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{99\cdot100}\)

\(A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{1}{4}-\frac{1}{100}\)

\(A=\frac{6}{25}\)

\(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{9900}\)

\(=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+.....+\frac{1}{99.100}\)

\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+....+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{4}-\frac{1}{100}\\ =\frac{24}{100}=\frac{6}{25}\)