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A = 1/1.2 + 1/3.4 + 1/5.6 + 1/7.8 + 1/9.10
A = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - 1/8 + 1/9 - 1/10
A = ( 1 + 1/3 + 1/5 + 1/7 + 1/9) - ( 1/2 + 1/4 + 1/6 + 1/8 + 1/10)
A = ( 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10) - 2.( 1/2 + 1/4 + 1/6 + 1/8 + 1/10)
A = ( 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10) - ( 1 + 1/2 + 1/3 + 1/4 + 1/5)
A = 1/6 + 1/7 + 1/8 + 1/9 + 1/10
B = 1/6.10 + 1/7.9 + 1/8.8 + 1/9.7 + 1/10.6
16B = 16/6.10 + 16/7.9 + 16/8.8 + 16/9.7 + 16/10.6
16B = 1/6 + 1/10 + 1/7 + 1/9 + 1/8 + 1/8 + 1/9 + 1/7 + 1/10 + 1/6
16B = 2.( 1/6 + 1/7 + 1/8 + 1/9 + 1/10)
8B = 1/6 + 1/7 + 1/8 + 1/9 + 1/10
Ta có A = 8B
=> A : B = 8
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(A=\frac{1}{2}-\frac{1}{10}\)
\(A=\frac{2}{5}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)-2\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)\(A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}-1-\frac{1}{2}-...-\frac{1}{1009}\)
\(A=\frac{1}{1010}+\frac{1}{2000}+...+\frac{1}{2018}\)
\(B=3028.\left(\frac{1}{1010.2018}+...+\frac{1}{2018.1010}\right)\)
\(B=\frac{3028}{1010.2018}+...+\frac{3028}{2018.1010}\)
\(B=\frac{1}{1010}+\frac{1}{2018}+...+\frac{1}{2018}+\frac{1}{1010}\)
\(B=2.\left(\frac{1}{1010}+...+\frac{1}{2018}\right)\)
\(=>\frac{A}{B}=\frac{1}{2}\)
\(A=\frac{1}{2.2}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
\(A=\frac{1}{4}+\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)
\(A=\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)
( gạch bỏ các phân số giống nhau)
\(A=\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(A=\frac{1}{4}+\frac{2}{9}\)
\(A=\frac{17}{36}\)
phần b, c bn lm tương tự như phần a nha
\(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{99\cdot100}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{50}\)
\(A=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)
a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{2003\cdot2004}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2003}-\frac{1}{2004}\)
\(=1-\frac{1}{2004}=\frac{2003}{2004}\)
b) Đặt A=\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2003\cdot2005}\)
\(2A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{1}{5\cdot7}+....+\frac{2}{2003\cdot2005}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2003}-\frac{1}{2005}\)
\(2A=1-\frac{1}{2005}\)
\(2A=\frac{2004}{2005}\)
\(A=\frac{2004}{2005}:2=\frac{2004}{2005}\cdot\frac{1}{2}=\frac{1002}{2005}\)
a)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2003.2004}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2003}-\frac{1}{2004}\)
\(=\frac{1}{1}-\frac{1}{2004}\)
\(\Rightarrow=\frac{2003}{2004}\)
b)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003+2005}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2003}-\frac{1}{2005}\)
\(=\frac{1}{1}-\frac{1}{2005}\)
\(\Rightarrow=\frac{2004}{2005}\)
a/ 1/1.3 + 1/3.4+ .... + 1/9.10
= 1–1/3+1/3–1/4+1/4–1/5+...+1/8–1/9+1/9–1/10
=1–1/10
=10/10–1/10
=9/10