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\(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{49.51}\)
\(=\frac{1}{2}.\left(\frac{2}{5}.7+\frac{2}{7}.9+\frac{2}{9}.11+...+\frac{2}{49}.51\right)\)
\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{51}\right)\)
\(=\frac{1}{2}.\frac{46}{255}\)
\(=\frac{23}{255}\)
\(\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{49.51}\)
\(\Rightarrow2 \left(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{49.51}\right)\)
\(\Rightarrow\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...+\frac{1}{49}-\frac{1}{51}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{51}=\frac{46}{255}\)
Vì biểu thức đã được nhân 2 nên giá trị của biểu thức là:
\(\frac{46}{255}:2=\frac{23}{255}\)
=\(\frac{1}{2}x\left(\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}+...+\frac{2}{2015x2017}\right)\)
=\(\frac{1}{2}x\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
=\(\frac{1}{2}x\left(\frac{1}{5}-\frac{1}{2017}\right)\)
=\(\frac{1}{2}x\frac{2012}{10085}\)
=\(\frac{1006}{10085}\)
Bạn gõ lại đề đi :v
Đọc chả hiểu đề gì cả ... đề k có x
Mà phía dưới có cái đáp số x= ... là sao ??
a)(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{11.12}\)). x=\(\frac{1}{3}\)
(1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{11}_{ }+\frac{1}{12}\)).x=\(\frac{1}{3}\)
(1+\(\frac{1}{12}\)).x=\(\frac{1}{3}\)
x=\(\frac{1}{3}:\frac{13}{12}\)
x=\(\frac{4}{13}\)
A)
\(\frac{1}{30}\)-\(\frac{1}{31}\)+\(\frac{1}{31}\)-\(\frac{1}{32}\)+\(\frac{1}{32}\)-\(\frac{1}{33}\)+...+\(\frac{1}{42}\)-\(\frac{1}{43}\)
=\(\frac{1}{30}\)-\(\frac{1}{43}\)
=\(\frac{13}{1290}\)
B)
=\(\frac{2}{2}\)X(\(\frac{1}{3.5}\)+\(\frac{1}{5.7}\)+\(\frac{1}{7.9}\)+\(\frac{1}{9.11}\))
=\(\frac{1}{2}\)X(\(\frac{2}{3.5}\)+\(\frac{2}{5.7}\)+\(\frac{2}{7.9}\)+\(\frac{2}{9.11}\))
=\(\frac{1}{2}\)X(\(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{9}\)+\(\frac{1}{9}\)-\(\frac{1}{11}\))
=\(\frac{1}{2}\)X(\(\frac{1}{3}\)-\(\frac{1}{11}\))
=\(\frac{1}{2}\)X\(\frac{8}{33}\)
=\(\frac{8}{66}\)=\(\frac{4}{33}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
Đặt A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
\(2A=\frac{1}{1}-\frac{1}{101}\)
\(2A=\frac{100}{101}\)
\(\Rightarrow A=\frac{100}{101}\div2\)
\(\Rightarrow A=\frac{50}{101}\)
1/5.7 + 1/7.9 + 1/9.11 + ... + 1/49.51
= 1/2 . (2/5.7 + 2/7.9 + 2/9.11 + ... + 2/49.51)
= 1/2 . (1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + ... + 1/49 - 1/51)
= 1/2 . (1/5 - 1/51)
= 1/2 . 46/255
= 23/255
S = \(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...\frac{1}{49}-\frac{1}{51}\)
S = \(\frac{1}{5}-\frac{1}{51}=\frac{46}{255}\)
Đặt \(A=\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{49.51}\)
\(A.2=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{49.51}\)
\(A.2=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{49}-\frac{1}{51}\)
\(A.2=\frac{1}{5}-\frac{1}{51}\)
\(A.2=\frac{46}{255}\)
\(A=\frac{23}{255}\)
Đặt \(A=\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{49.51}\)
\(2A=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{49.51}\)
\(2A=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{49}-\frac{1}{51}\)
\(2A=\frac{1}{5}-\frac{1}{51}\)
\(A=\frac{46}{255}.\frac{1}{2}=\frac{23}{255}\)
Đặt \(A=\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{49.51}\)
\(A.2=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{49.51}\)
\(A.2=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{49}-\frac{1}{51}\)
\(A.2=\frac{1}{5}-\frac{1}{51}\)
\(A.2=\frac{46}{255}\)
\(A=\frac{23}{255}\)
= 1 / 5 - 1 / 7 + 1 / 7 - 1 / 9 + 1 / 9 - 1 /11 + .....+ 1 / 49 - 1 /51
= 1 / 5 - 1/ 51
= 46 / 255
\(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{49.50}\)
=\(\frac{1}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{49.50}\right)\)
=\(\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{49}-\frac{1}{50}\right)\)
=\(\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{50}\right)\)
=\(\frac{1}{2}.\left(\frac{10}{50}-\frac{1}{50}\right)\)
=\(\frac{1}{2}.\frac{49}{50}\)
=\(\frac{49}{100}\)
k nha
\(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+....+\frac{1}{49.51}\)
\(=\frac{1}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{49.51}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{51}\right)\)
\(=\frac{1}{2}.\frac{46}{255}=\frac{23}{255}\)
\(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{49.51}\)
\(=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{51}\right)=\frac{1}{2}.\frac{46}{255}=\frac{23}{255}\)
Vậy tổng trên bằng \(\frac{23}{255}\)
K cho mk nha, mk làm nhanh nhất nhá
\(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{49.51}\)
\(=\frac{1}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+....+\frac{2}{49.51}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{51}\right)\)
\(=\frac{1}{2}.\frac{46}{255}\)
\(=\frac{23}{255}\)