Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt tông trên là A
\(\dfrac{2A}{7}=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{2023-2021}{2021.2023}=\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2023}=1-\dfrac{1}{2023}=\dfrac{2022}{2023}\)
\(\Rightarrow A=\dfrac{7.2022}{2.2023}=\dfrac{1011}{289}\)
\(M=\frac{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{11}}=\frac{3\left(\frac{1}{5}+\frac{1}{7}-\frac{3}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}=\frac{3}{4}\) \(\frac{3}{4}\) \(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}=2-\frac{2}{101}=\frac{200}{101}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(B=2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{101}\right)\)
\(B=2.\frac{100}{101}=\frac{200}{101}\)
Ta có :
M= \(\dfrac{3+3-3+\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{4+4-4+\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\)= \(\dfrac{3+3-3}{4+4-4}=\dfrac{3}{4}\)
b) Nhận xét thấy: \(\dfrac{2}{1.3}=1-\dfrac{1}{3};\dfrac{1}{3.5}=\dfrac{1}{3}-\dfrac{1}{5};...\)
Ta có:
B= 1-\(\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
B= 1- \(\dfrac{1}{101}\)= \(\dfrac{100}{101}\)
Vậy B= \(\dfrac{100}{101}\)
= \(\dfrac{5}{2}(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2019}-\dfrac{1}{2021})\)
= \(\dfrac{5}{2}\left(1-\dfrac{1}{101}\right)\)
= \(\dfrac{5}{2}.\dfrac{100}{101}\)
= \(\dfrac{250}{101}\)
A = \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
A=\(\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{50}{100}-\dfrac{1}{100}=\dfrac{49}{100}\)
B = \(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{49.51}\)
B = \(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{49}-\dfrac{1}{51}\)
B = \(\dfrac{1}{2}-\dfrac{1}{51}=\dfrac{51}{102}-\dfrac{2}{102}=\dfrac{49}{102}\)
\(\dfrac{6}{1.3}+\dfrac{6}{3.5}+...+\dfrac{6}{99.100}\\ =3\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.100}\right)\\ =3\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =3\left(1-\dfrac{1}{100}\right)\\ =3.\dfrac{99}{100}\\ =\dfrac{297}{100}\)
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)=\dfrac{1}{2}.\dfrac{100}{101}=\dfrac{50}{101}\)
\(\dfrac{5}{1\cdot3}+\dfrac{5}{3\cdot5}+\dfrac{5}{5\cdot7}+...+\dfrac{5}{201\cdot203}\)
= \(\dfrac{5}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{201\cdot203}\right)\)
= \(\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{201}-\dfrac{1}{203}\right)\)
= \(\dfrac{5}{2}\left(1-\dfrac{1}{203}\right)\)
= \(\dfrac{5}{2}\cdot\dfrac{202}{203}=\dfrac{505}{203}\)
Ta có :
\(\dfrac{5}{1.3}+\dfrac{5}{3.5}+\dfrac{5}{5.7}+...+\dfrac{5}{201.203}\)
\(=\dfrac{5}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{201.203}\right)\)
\(=\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-...+\dfrac{1}{201}-\dfrac{1}{203}\right)\)
\(=\dfrac{5}{2}\left(1-\dfrac{1}{203}\right)\)
\(=\dfrac{5}{2}.\dfrac{202}{203}\)
\(=\dfrac{505}{203}\)
\(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\) + ..... + \(\dfrac{2}{95.97}\)
= 1 - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + .... + \(\dfrac{1}{95}\) - \(\dfrac{1}{97}\)
= \(1-\dfrac{1}{97}\)
= \(\dfrac{96}{97}\)
\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+...+\dfrac{2}{95\times97}\)
\(=\dfrac{2}{3}\left(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{95\times97}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)
\(=\dfrac{2}{3}\left(1-\dfrac{1}{97}\right)\)\(=\dfrac{2}{3}\times\dfrac{96}{97}\)\(=\dfrac{64}{97}\)
Đặt tổng trên là A
\(\dfrac{2A}{7}=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2021.2023}=\)
\(=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{2023-2021}{2021.2023}=\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2023}=\)
\(=\dfrac{2022}{2023}\Rightarrow A=\dfrac{7.2022}{2.2023}\)
\(A=\dfrac{7}{1.3}+\dfrac{7}{3.5}+\dfrac{7}{5.7}+...+\dfrac{7}{2021.2023}\\ \Rightarrow\dfrac{2}{7}A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2021.2023}\\ \Rightarrow\dfrac{2}{7}A=2-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{2}{7}+...+\dfrac{2}{2021}-\dfrac{2}{2023}\\ \Rightarrow\dfrac{2}{7}A=2-\dfrac{2}{2023}=\dfrac{4044}{2023}\Rightarrow A=\dfrac{2022}{289}\)