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\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
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 = \(\frac{1}{5}-\frac{1}{51}=\frac{46}{255}\)
\(C=\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{37\cdot39}\)
\(2C=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{37\cdot39}\)
\(2C=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{39}\)
\(2C=\dfrac{1}{3}-\dfrac{1}{39}\)
\(2C=\dfrac{4}{13}\)
\(C=\dfrac{2}{13}\)
\(C=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{37.39}\)
\(C=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{39}\right)\)
\(C=\frac{1}{2}.\frac{12}{39}\)
\(C=\frac{4}{26}=\frac{2}{13}\)
=\(\frac{1}{2}x\left(\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}+...+\frac{2}{2015x2017}\right)\)
=\(\frac{1}{2}x\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
=\(\frac{1}{2}x\left(\frac{1}{5}-\frac{1}{2017}\right)\)
=\(\frac{1}{2}x\frac{2012}{10085}\)
=\(\frac{1006}{10085}\)
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\(\frac{8}{33}\)
=\(\frac{8}{66}\)=\(\frac{4}{33}\)
đặt A=tổng trên
\(2A=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{28}-\frac{1}{30}\)
\(2A=\frac{1}{5}-\frac{1}{30}\)
\(A=\frac{1}{6}:2\)
\(A=\frac{1}{12}\)
la 1/12