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\(\Rightarrow A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(\Rightarrow A=\frac{1}{2}\left(1-\frac{1}{2015}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{2014}{2015}=\frac{1007}{2015}\)
A = 1/1.3 + 1/3.5 + 1/5.7 +........+ 1/1999.2001
2.A = 2/1.3 + 2/3.5 + 2/5.7 +........+ 2/1999.2001
2.A = 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/1999 - 1/2001
2.A = 1 - 1/2001
2.A = 2000/2001
Vậy A =1000/2001
B = 1/3.5 + 1/5.7 + 1/7.9 +........+ 1/99.101
2.A = 2/3.5 + 2/5.7 + 2/7.9 +........+ 2/99.101
2.A = 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/99 - 1/101
2.A = 1/3 - 1/101 = 98/303
Vậy A =49/303
\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{1999.2001}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{1999.2001}\)
\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{1999}-\frac{1}{2001}\)
\(2A=\frac{1}{1}-\frac{1}{2001}=\frac{2000}{2001}\)
\(A=\frac{2000}{2001}.\frac{1}{2}=\frac{1000}{2001}\)
a, \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)
=2.(\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\))
=\(2.\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
=\(\frac{2}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{100}{101}\)
b, \(\frac{5}{1.3}+\frac{5}{3.5}+...+\frac{5}{99.101}\)
=\(5.\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\right)\)
=\(5.\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)
=\(\frac{250}{101}\)
\(=\frac{5}{2}.\frac{100}{101}\)
a,21.321.3+23.523.5+25.725.7+....+299.101
=>\(\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\)
=>\(\frac{1}{1}-\frac{1}{101}\)
=>\(\frac{100}{101}\)
b,
51.351.3+53.553.5+55.755.7+....+599.101
=>\(\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{99.101}\right)\)
=>\(\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\right)\)
=>\(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{101}\right)\)
=>\(\frac{5}{2}.\frac{100}{101}\)
=>\(\frac{250}{101}\)
\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{47}-\frac{1}{49}\right)=\frac{1}{x}\)
\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{49}\right)=\frac{1}{x}\Rightarrow x=\frac{49}{24}\)
\(\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+...+\frac{1}{2}.\left(\frac{1}{47}-\frac{1}{49}\right)=\frac{1}{x}\)
\(\frac{1}{2}.\left(1-\frac{1}{49}\right)=\frac{1}{x}\)
\(\frac{24}{49}=\frac{1}{x}\)\(\Rightarrow x=\frac{49}{24}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
Đặt A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
\(2A=\frac{1}{1}-\frac{1}{101}\)
\(2A=\frac{100}{101}\)
\(\Rightarrow A=\frac{100}{101}\div2\)
\(\Rightarrow A=\frac{50}{101}\)
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{x(x+2)}=\frac{20}{41}\)
\(\Rightarrow\frac{1}{2}\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{x(x+2)}\right]=\frac{20}{41}\)
\(\Rightarrow\frac{1}{2}\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right]=\frac{20}{41}\)
\(\Rightarrow\frac{1}{2}\left[1-\frac{1}{x+2}\right]=\frac{20}{41}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{20}{41}:\frac{1}{2}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{40}{41}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{40}{41}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{41}\Leftrightarrow x+2=41\Leftrightarrow x=39\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{20}{41}.\)
\(1-\frac{1}{x+2}=\frac{20}{41}\Rightarrow\frac{1}{x+2}=\frac{21}{41}=\frac{21}{21x+42}\Rightarrow21x+42=41\Rightarrow x=-\frac{1}{21}\)
2/7A=2/1.3+2/3.5+...+2/99+101
2/7A=1-1/3+1/3-1/5+...+1/99-1/101
2/7A=1-1/101
2/7A=100/101
A=350/101
B=(1+1+1+1)-(1/2+1/6+1/12+1/20)
=4-(1/1.2+1/2.3+1/3.4+1/4.5)
=4-(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5)
=4-(1-1/5)
=4-4/5
=16/5