Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
\(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{990}\)
\(=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)
\(=\frac{1}{4}-\frac{1}{100}\)
\(=\frac{6}{25}\)
Trần Thùy Dung nó đã bảo \(990\ne99\cdot100\) rùi mà vẫn tách như v
=\(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
=\(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
=\(\frac{1}{4}-\frac{1}{100}\)
=\(\frac{24}{100}=\frac{6}{25}\)
\(B1\)
\(=\frac{1}{1}-\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{37}-\frac{1}{38}-\frac{1}{39}\)
\(=1-\frac{1}{39}\)
\(=\frac{38}{39}\)
\(B2\)
\(=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+.....+\frac{1}{99\cdot100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+......+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}\)
\(=\frac{25}{100}-\frac{1}{100}\)
\(=\frac{24}{100}\)
\(=\frac{6}{25}\)
Bài 1 :
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\)
\(=\frac{1}{1.2}-\frac{1}{38.39}\)
\(=\frac{370}{741}\)
\(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)
\(=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{3}-\frac{1}{8}=\frac{5}{24}\)
\(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)
\(=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{3}-\frac{1}{8}\)
\(=\frac{8-3}{24}=\frac{5}{24}\)
\(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{9900}\)
\(A=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{99\cdot100}\)
\(A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{4}-\frac{1}{100}\)
\(A=\frac{6}{25}\)
\(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{9900}\)
\(=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+.....+\frac{1}{99.100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+....+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}\\ =\frac{24}{100}=\frac{6}{25}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
1/20 + 1/30 + 1/42 + ... + 1/9900
= 1/4.5 + 1/5.6 + 1/6.7 + ... + 1/99.100
= 1/4 - 1/5 + 1/5 - 1/6 + ... + 1/99 - 1/100
= 1/4 - 1/100
= 6/25
bn ơi,hình nhưa sai đề,số 990 mik ko phân tích ra tích của 2 số tự nhiên liên tiếp được,chắc là sai đề nha bn,bn kiểm tra lại đề rồi đăng câu hỏi khác nhé!!!,để mik xem lại coi.....
kiểm tra nha,nếu mà đúng đề thì để mik xem lại...
T.T