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\(=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}\)
\(=\frac{24}{100}=\frac{6}{25}\)
\(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{9900}\)
\(\Rightarrow A=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{99.100}\)
\(\Rightarrow A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A=\frac{1}{4}-\frac{1}{100}\)
\(\Rightarrow A=\frac{25}{100}-\frac{1}{100}\)
\(\Rightarrow A=\frac{24}{100}\)
\(\Rightarrow A=\frac{6}{25}\)
\(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{9900}\)
\(A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}-\frac{1}{6}+\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{4}-\frac{1}{100}\)
\(A=\frac{6}{25}\)
\(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{9900}\)
\(A=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{99.100}\)
\(A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{4}-\frac{1}{100}=\frac{6}{25}\)
Vậy A=6/25
\(A=\frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}+.....+\frac{1}{99\times100}\)
\(A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+.....+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{4}-\frac{1}{100}\)
\(A=\frac{24}{100}=\frac{6}{25}\)
\(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{9900}\)
\(A=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{99\cdot100}\)
\(A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{4}-\frac{1}{100}=\frac{25}{100}-\frac{1}{100}=\frac{24}{100}=\frac{6}{25}\)
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 + ... + 1/9900
= 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
\(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)
\(=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{3}-\frac{1}{8}=\frac{5}{24}\)
\(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)
\(=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{3}-\frac{1}{8}\)
\(=\frac{8-3}{24}=\frac{5}{24}\)
\(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{9900}\)
\(A=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{99\cdot100}\)
\(A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{4}-\frac{1}{100}\)
\(A=\frac{6}{25}\)
\(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{9900}\)
\(=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+.....+\frac{1}{99.100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+....+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}\\ =\frac{24}{100}=\frac{6}{25}\)