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\(C=\frac{2}{4.7}-\frac{3}{5.9}+\frac{2}{7.10}-\frac{3}{9.13}+...+\frac{2}{301.304}-\frac{3}{401.405}\)
\(C=\left(\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{301.304}\right)-\left(\frac{3}{5.9}+\frac{3}{9.13}+...+\frac{3}{401.405}\right)\)
\(C=\frac{2}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{301.304}\right)-\frac{3}{4}\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{401.405}\right)\)
\(C=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{301}-\frac{1}{304}\right)-\frac{3}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+..+\frac{1}{401}-\frac{1}{405}\right)\) \(C=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{304}\right)-\frac{3}{4}\left(\frac{1}{5}-\frac{1}{405}\right)\)
\(C=\frac{25}{152}-\frac{4}{27}\)
\(C=\frac{67}{4104}\)
1.
a.
\(\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{7}\right)\)
\(=\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\)
\(=\frac{35-21-15}{105}\)
\(=-\frac{1}{105}\)
b.
\(\frac{3}{5}-\left(\frac{3}{4}-\frac{1}{2}\right)\)
\(=\frac{3}{5}-\frac{3}{4}+\frac{1}{2}\)
\(=\frac{12-15+10}{20}\)
\(=\frac{7}{20}\)
c.
\(\frac{4}{7}-\left(\frac{2}{5}+\frac{1}{3}\right)\)
\(=\frac{4}{7}-\frac{2}{5}-\frac{1}{3}\)
\(=\frac{60-42-35}{105}\)
\(=-\frac{17}{105}\)
2.
a.
\(S=-\frac{1}{1\times2}-\frac{1}{2\times3}-\frac{1}{3\times4}-...-\frac{1}{\left(n-1\right)\times n}\)
\(S=-\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{\left(n-1\right)\times n}\right)\)
\(S=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)
\(S=-\left(1-\frac{1}{n}\right)\)
\(S=-1+\frac{1}{n}\)
b.
\(S=-\frac{4}{1\times5}-\frac{4}{5\times9}-\frac{4}{9\times13}-...-\frac{4}{\left(n-4\right)\times n}\)
\(S=-\left(\frac{4}{1\times5}+\frac{4}{5\times9}+\frac{4}{9\times13}+...+\frac{4}{\left(n-4\right)\times n}\right)\)
\(S=-\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{n-4}-\frac{1}{n}\right)\)
\(S=-\left(1-\frac{1}{n}\right)\)
\(S=-1+\frac{1}{n}\)
Chúc bạn học tốt
Bài 1 :
\(A=\frac{1}{3}-\frac{3}{4}-\frac{\left(-3\right)}{5}+\frac{1}{72}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)
\(\Rightarrow A=\frac{3}{9}-\frac{3}{4}+\frac{9}{15}+\frac{1}{72}-\frac{2}{9}-\frac{2}{72}+\frac{1}{15}\)
\(\Rightarrow A=\left(\frac{3}{9}-\frac{2}{9}\right)+\left(\frac{9}{15}+\frac{1}{15}\right)+\left(\frac{1}{72}+\frac{-2}{72}\right)-\frac{3}{4}\)
\(\Rightarrow A=\frac{1}{9}+\frac{2}{3}+\frac{-1}{72}-\frac{3}{4}=\frac{8}{72}+\frac{48}{72}+\frac{-1}{72}-\frac{54}{72}\)
\(\Rightarrow A=\frac{1}{72}\)
Vậy : \(A=\frac{1}{72}\)
Bài 2:
Bạn tham khảo tại đây nhé: Câu hỏi của Linh Nguyễn
Chúc bạn học tốt!
Ta có :
\(M=-\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-...-\frac{4}{\left(n+4\right)n}\)
\(\Leftrightarrow\)\(M=-\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{n+4}-\frac{1}{n}\right)\)
\(\Leftrightarrow\)\(M=-\left(1-\frac{1}{n}\right)\)
\(\Leftrightarrow\)\(M=-\frac{n}{n}+\frac{1}{n}\)
\(\Leftrightarrow\)\(M=\frac{-n+1}{n}\)
Vậy \(M=\frac{-n+1}{n}\)
\(\frac{1}{2}+\frac{2}{3}-\frac{3}{4}+\frac{4}{5}-\frac{5}{6}+\frac{6}{7}+\frac{5}{6}-\frac{4}{5}+\frac{3}{4}-\frac{2}{3}+\frac{1}{2}\)
\(=\left(\frac{1}{2}+\frac{1}{2}+\frac{6}{7}\right)+\left(\frac{2}{3}-\frac{2}{3}\right)+\left(\frac{-3}{4}+\frac{3}{4}\right)+\left(\frac{4}{5}-\frac{4}{5}\right)+\left(\frac{-5}{6}+\frac{5}{6}\right)\)
\(=\frac{13}{7}+0+0+0+0\)
\(=\frac{13}{7}\)
\(\frac{1}{2}+\frac{2}{3}-\frac{3}{4}+\frac{4}{5}-\frac{5}{6}+\frac{6}{7}+\frac{5}{6}-\frac{4}{5}+\frac{3}{4}-\frac{2}{3}+\frac{1}{2}.\)
\(=\left(\frac{1}{2}+\frac{1}{2}\right)+\left(\frac{2}{3}-\frac{2}{3}\right)-\left(\frac{3}{4}-\frac{3}{4}\right)+\left(\frac{4}{5}-\frac{4}{5}\right)-\left(\frac{5}{6}-\frac{5}{6}\right)+\frac{6}{7}\)
\(=1+0-0+0+\frac{6}{7}\)
\(=1+\frac{6}{7}=1\frac{6}{7}\)
a) A = \(\frac{5}{1.4}+\frac{29}{4.7}+\frac{71}{7.10}+....+\frac{10301}{100.103}\) (có 34 số hạng)
A = \(\frac{4+1}{1.4}+\frac{4.7+1}{4.7}+\frac{7.10+1}{7.10}+....+\frac{100.103+1}{103.100}\)
A = \(1+\frac{1}{1.4}+1+\frac{1}{4.7}+1+\frac{1}{7.10}+....+1+\frac{1}{100.103}\)
A = \(1.34+\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)\)
A = \(34+\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)
A = \(34+\frac{1}{3}.\left(1-\frac{1}{103}\right)\)
A = \(34+\frac{1}{3}\cdot\frac{102}{103}\)
A = \(34+\frac{34}{103}=\frac{3536}{103}\)
1 Tính :
a) \(A=\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{\left(n-1\right).n}\)
\(=\frac{1}{1.2}-\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{n}\right)\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{n}\)
\(=\frac{1}{n}\)
b) \(B=\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-...-\frac{4}{\left(n-4\right).n}\)
\(=\frac{4}{1.5}-\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{\left(n-4\right).n}\right)\)
\(=\frac{4}{5}-\left(\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{\left(n-4\right).n}\right)\)
\(=\frac{4}{5}-\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{n-4}-\frac{1}{n}\right)\)
\(=\frac{4}{5}-\left(\frac{1}{5}-\frac{1}{n}\right)\)
\(=\frac{4}{5}-\frac{1}{5}+\frac{1}{n}\)
\(=\frac{3}{5}+\frac{1}{n}\)
c) \(C=1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{10}}\)
\(=1-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
Đặt \(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(\Rightarrow C=1-B\left(1\right)\)
\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
Lấy 2B trừ B ta có :
\(2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(B=1-\frac{1}{2^{10}}\left(2\right)\)
Thay (2) vào (1) ta có :
\(C=1-\left(1-\frac{1}{10}\right)\)
\(=1-1+\frac{1}{10}\)
\(=\frac{1}{10}\)
Vậy \(C=\frac{1}{10}\)