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a) \(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1-n}{n.\left(n+1\right)}=\frac{1}{n.\left(n+1\right)}\)
\(\frac{1}{n}.\frac{1}{n+1}=\frac{1}{n.\left(n+1\right)}\)
vậy \(\frac{1}{n}và\frac{1}{n+1}\)có hiệu và tích bằng nhau
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{9}\)
do có các cặp âm và dương nên gạch vậy A=\(\frac{1}{2}-\frac{1}{9}\)=\(\frac{7}{18}\)
B=\(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{10.11}\)
cách lm tương tự câu A
vậy B= \(\frac{1}{4}-\frac{1}{11}\)=\(\frac{7}{44}\)
\(a)\)\(\frac{1}{n}\cdot\frac{1}{n+1}=\frac{1}{n(n+1)}\) ; \(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1-n}{n(n+1)}=\frac{1}{n(n+1)}\)
\(b)A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(A=\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+\frac{1}{11\cdot12}\)
\(=(\frac{1}{5}-\frac{1}{6})+(\frac{1}{6}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{8})+(\frac{1}{8}-\frac{1}{9})+(\frac{1}{9}-\frac{1}{10})+(\frac{1}{10}-\frac{1}{11})+(\frac{1}{11}-\frac{1}{12})\)
\(=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
a) Ta có hiệu của chúng là:
\(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\left(1\right)\)
Mặt khác, ta lại có tích của chúng là:
\(\frac{1}{n}.\frac{1}{n+1}=\frac{1}{n\left(n+1\right)}\left(2\right)\)
Từ (1) và (2) suy ra: \(\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n}.\frac{1}{n+1}\)
Vậy tích của hai phân số này bằng hiệu của chúng (hiệu của phân số lớn trừ phân số nhỏ)
b) \(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+....+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
b)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{2}-\frac{1}{100}\)
\(A=\frac{49}{100}\)
\(B=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
\(B=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{11}-\frac{1}{12}\)
\(B=\frac{1}{5}-\frac{1}{12}\)
\(B=\frac{7}{60}\)
a) Ta có:
\(\frac{1}{n}.\frac{1}{n+1}=\frac{1}{n\left(n+1\right)}\) ; \(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{1\left(n+1\right)}\)
Vậy \(\frac{1}{n}.\frac{1}{n+1}=\frac{1}{n}-\frac{1}{n+1}\)
b) \(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(A=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+....+\frac{100-99}{99.100}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{2}-\frac{1}{100}\)
\(A=\frac{49}{100}\)
\(B=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(B=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
\(B=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}-\frac{1}{11}-\frac{1}{12}\)
\(B=\frac{1}{5}-\frac{1}{12}\)
\(B=\frac{7}{60}\)
http://olm.vn/hỏi-đáp/question/584545.html chờ xí tui thấy cái tên rồi giải cho bài 2
a,\(\frac{2}{1.3}+...\frac{2}{99.101}\)
\(=\frac{3-1}{1.3}+...+\frac{101-99}{99.101}\)
\(=\frac{3}{1.3}-\frac{1}{1.3}+...+\frac{101}{99.101}-\frac{99}{99.101}\)
\(=\frac{1}{1}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{1}-\frac{1}{101}\)
\(\frac{100}{101}\)
Ta có : \(\frac{\frac{3}{5}+\frac{3}{7}-\frac{1}{3}+\frac{3}{11}}{\frac{6}{5}+\frac{6}{7}-\frac{2}{3}+\frac{6}{11}}=\frac{\frac{3}{5}+\frac{3}{7}-\frac{1}{3}+\frac{3}{11}}{2\left(\frac{3}{5}+\frac{3}{7}-\frac{1}{3}+\frac{3}{11}\right)}=\frac{1}{2}\)
Lại có : \(\frac{\left(\frac{1}{4}-\frac{1}{5}-\frac{1}{20}\right).2021}{\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}}=\frac{0.2021}{\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}}=0\)
Khi đó \(B=\frac{1}{2}+0=\frac{1}{2}\)
a)\(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n-1\right)}=\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\)
\(\Rightarrow\frac{1}{n\left(n+1\right)}=\frac{1}{n}.\frac{1}{n+1}\)
b) \(C=\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}+\frac{1}{6}.\frac{1}{7}+\frac{1}{7}.\frac{1}{8}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}+0+0+0+0+0-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}=\frac{4}{8}-\frac{1}{8}=\frac{4-1}{8}=\frac{3}{8}\)