Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(A=\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+...+\frac{1}{25\cdot27\cdot29}\)
\(\Rightarrow4A=\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+...+\frac{4}{25\cdot27\cdot29}\)
\(\Rightarrow4A=\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+...+\frac{1}{25\cdot27}-\frac{1}{27\cdot29}\)
\(\Rightarrow4A=\frac{1}{1\cdot3}-\frac{1}{27\cdot29}=\frac{1}{3}-\frac{1}{783}=\frac{261}{783}-\frac{1}{783}=\frac{260}{783}\)
\(\Rightarrow A=\frac{\frac{260}{783}}{4}=\frac{65}{783}\)
b) \(\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\)
\(\Rightarrow100\cdot\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=100\cdot\left(\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\right)\)
\(\Rightarrow\left(\frac{100}{1\cdot101}+\frac{100}{2\cdot102}+...+\frac{100}{10\cdot110}\right)x=10\cdot\left(\frac{10}{1\cdot11}+\frac{10}{2\cdot12}+...+\frac{10}{100\cdot110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{10}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)\)
\(\Rightarrow x=10\cdot\)
Gọi tử số là \(C\)và mẫu số là \(D\)
Ta có:
\(A=\frac{C}{D}\)
\(C=\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.102}+...+\frac{1}{101.400}\)
\(C=\frac{1}{299}\left[\left(1-\frac{1}{300}\right)\right]+\left(\frac{1}{2}-\frac{1}{301}\right)+\left(\frac{1}{3}-\frac{1}{302}\right)+...+\left(\frac{1}{101}-\frac{1}{400}\right)\)
\(C=\frac{1}{299}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-\frac{1}{302}-...-\frac{1}{400}\right)\)
\(D=\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{299.400}\)
\(D=\frac{1}{101}\left[\left(1-\frac{1}{102}\right)+\left(\frac{1}{2}-\frac{1}{103}\right)+\left(\frac{1}{3}-\frac{1}{104}\right)+...+\left(\frac{1}{299}-\frac{1}{400}\right)\right]\)
\(D=\frac{1}{101}\left(1+\frac{1}{2}+\frac{1}{3}+..+\frac{1}{299}-\frac{1}{102}-\frac{1}{103}-\frac{1}{104}-...-\frac{1}{400}\right)\)
\(D=\frac{1}{101}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-\frac{1}{302}-...-\frac{1}{400}\right)\)
\(\Rightarrow A=\frac{C}{D}=\frac{\frac{1}{299}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-\frac{1}{302}-...-\frac{1}{400}\right)}{\frac{1}{101}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-\frac{1}{302}-...-\frac{1}{400}\right)}\)
\(=\frac{\frac{1}{299}}{\frac{1}{101}}=\frac{101}{299}.\)
Vậy \(A=\frac{101}{299}.\)
A=2/1.3 + 2/3.5 + 2/5.7 + ... + 2/99.101
A= 2 - 1/3 + 1/3 - 1/5 + 1/5 - ... + 2/99 - 2/101
A = 2 - 2/101 = 200/101
B = 3-1/3+1/3-1/5+1/5-...+3/49-3/51
B = 3-3/51(tự tính nhé)
C = 5(5/1.6+5/6.11+5/11.16+....+5/26-5/31
C = 5(5-1/31)(tự tính)
D rút gon cho 2 rồi 3D , sau đó 5(3/.... tương tự các cách làm trên)
2E nhân lên rồi giải giống trên
3F Rồi nhân 4/77 và rút gọn thì tính được
a, A= \(\frac{1}{1}\)- \(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+......+\(\frac{1}{99}\)-\(\frac{1}{100}\)
A=\(\frac{1}{1}\)-\(\frac{1}{100}\)+(-\(\frac{1}{3}\)+\(\frac{1}{3}\)-.....-\(\frac{1}{99}\)+\(\frac{1}{99}\))
A=\(\frac{1}{1}\)-\(\frac{1}{100}\)+0
A=1-\(\frac{1}{100}\)=\(\frac{100}{100}\)-\(\frac{1}{100}\)=\(\frac{99}{100}\)
\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(=7.\frac{1}{10.11}+7.\frac{1}{11.12}+7.\frac{1}{12.13}+...+7.\frac{1}{69.70}\)
\(=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}=\frac{3}{5}\)
\(A=7.\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(=\frac{6}{70}\)
\(=\frac{3}{35}\)
