Ta có:\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)=\frac{1}{2}\left(1-\frac{1}{21}\right)=\frac{1}{2}.\frac{20}{21}=\frac{10}{21}\)

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}\)\(+...+\frac{1}{19.21}\)

=\(\frac{2}{2}\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}\right)\)

=\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{19.21}\right)\)

=\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{19}-\frac{1}{21}\right)\)

=\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{21}\right)\)

=\(\frac{1}{2}.\frac{20}{21}\)

=\(\frac{20}{42}=\frac{10}{21}\)

Đặt :

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}\)

\(\Leftrightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{19.21}\)

\(\Leftrightarrow2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{21}\)

\(\Leftrightarrow2A=1-\frac{1}{21}\)

\(\Leftrightarrow2A=\frac{20}{21}\)

\(\Leftrightarrow A=\frac{10}{21}\)

\(2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.100}\right)\)

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.100}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

2*(1/1*3+1/3*5+.......+1/99*100)

=2*(2/1*3+2/3*5+.....+2/99*100)*1/2

=1/3-1/5+1/5-1/7+....+1/99-1/100

=1/3-1/100

=100/300-3/300

=97/300

2(1/1.3+1/3.5+1/5.7+...+1/99.100)

=2/1.3+2/3.5+2/5.7+...+2/99.100

=1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/100

=1-1/100

=99/100

Ta có:

\(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+\frac{1}{4.6}+\frac{1}{5.7}+\frac{1}{6.8}+\frac{1}{7.9}+\frac{1}{8.10}\)

\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{8}-\frac{1}{10}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}....+\frac{1}{7}-\frac{1}{9}\right)+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{10}\right)\)

\(=\frac{1}{2}.\frac{8}{9}+\frac{1}{2}.\frac{2}{5}=\frac{1}{2}.\left(\frac{8}{9}+\frac{2}{5}\right)=\frac{1}{2}.\frac{58}{45}=\frac{29}{45}\)

29/45 bạn ạ

bằng 29/45 k cho em với

Đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{99\cdot101}\)

\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{99}-\frac{1}{101}\)

\(2A=\frac{100}{101}\)

\(A=\frac{50}{101}\)

b) \(\frac{2^{10}+3^{31}+2^{40}+3^6}{2^{11}\cdot3^{31}+2^{41}\cdot3^6}=\frac{2^{10}+2^{40}}{2^{11}+2^{41}}\)

\(\frac{2^{10}+2^{40}}{2^{11}+2^{41}}=\frac{1}{2}\)

=1/2x(1/1.3+1/3.5+...+1/99.101)

=1/2.(1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101)

=1/2.(1-1/101)

=1/2.100/101

=50/101

chúc bạn học tốt

Q = \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2013.2015}\)

Q = \(\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{1}{2013.2015}\right)\)

Q =  \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2015}\right)\)

Q = \(\frac{1}{2}.\frac{2012}{6045}=\frac{1002}{6045}\)

\(Q=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2013.2015}\)

\(\Rightarrow Q.2=2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2013.2015}\right)\)

\(\Rightarrow Q.2=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{2013.2015}\)

\(\Rightarrow Q.2=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2013}-\frac{1}{2015}\)

\(\Rightarrow Q.2=\frac{1}{3}-\frac{1}{2015}\)

\(\Rightarrow Q.2=\frac{2012}{6045}\)

\(\Rightarrow Q=\frac{2012}{6045}.\frac{1}{2}=\frac{1006}{6045}\)

Mk tinh nhẩm, nên ko bt kết quả có đúng ko

nên bn thử tính lại kết quả nha!!!

Chúc bn hok tốt!!!

có công thư rồi mà bài này dễ đợi mk 3' nhé mk giải cho

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2015}\right)=\frac{1}{2}.\frac{2012}{6045}=\frac{1006}{6045}\)

lm tắt cu~g chả bt đúng ko ^^, thông cảm 

\(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)

\(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2015}\right)\)

\(\frac{1}{2}.\frac{2012}{6045}\)

=\(\frac{1006}{6045}\)  

ok bạn nhé

\(-\frac{2}{1.3}-\frac{2}{3.5}-\frac{2}{5.7}-\frac{2}{7.9}-\frac{2}{9.11}-\frac{2}{11.13}-\frac{2}{13.15}\)

\(=\left(-\frac{2}{1.3}\right)+\left(-\frac{2}{3.5}\right)+\left(-\frac{2}{5.7}\right)+\left(-\frac{2}{7.9}\right)+\left(-\frac{2}{9.11}\right)+\left(-\frac{2}{11.13}\right)+\left(-\frac{2}{13.15}\right)\)

\(=\left(-2\right).\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\right)\)

\(=\left(-2\right).\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)

\(=\left(-2\right).\left(1-\frac{1}{15}\right)=\left(-2\right).\frac{14}{15}\)

\(=-\frac{28}{15}\)

tks bạn nha ^_^ Miu Ti