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S = \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{820}\)
S = \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{1640}\)
S = \(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{40.41}\right)\)
S = \(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{40}-\frac{1}{41}\right)\)
S = \(2.\left(\frac{1}{2}-\frac{1}{41}\right)\)
S = \(2.\frac{39}{82}\)
S = \(\frac{39}{41}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{97}-\frac{1}{97}\right)-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
~ Hok tốt ~
\(\)
S = 1/3.5 + 1/5.7 + 1/7.9 +...+1/97.99
S = 1 - ( 1/3+1/3-1/5+1/5-+1/7+1/7+...+1/97-1/99)
S = 1 - 1/99
S = 98/99
\(2S=\frac{2}{3.5}+\frac{2}{5.7}+.......+\frac{2}{97.99}\)
\(2S=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{99}\)
\(2S=\frac{1}{3}-\frac{1}{99}\)
\(2S=\frac{33}{99}-\frac{1}{99}\)
\(S=\frac{32}{99}:2\)
\(S=\frac{16}{99}\)
\(S=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\)
\(2S=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)
\(2S=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
\(2S=\dfrac{1}{3}-\dfrac{1}{99}\)
\(2S=\dfrac{32}{99}\)
\(S=\dfrac{32}{99}:2\)
\(S=\dfrac{16}{99}\)
Cảm ơn bạn rất nhiều! Bạn đã cứu mình rùi! Xin trân thành cảm ơn!
= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
\(=\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{32}{99}\)
Ta có : \(S=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(\Rightarrow2S=2\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\right)\)
\(\Rightarrow2S=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
\(\Rightarrow2S=\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+...+\left(\frac{1}{97}-\frac{1}{99}\right)\)
\(\Rightarrow2S=\left(\frac{1}{3}-\frac{1}{99}\right)+\left[\left(\frac{1}{5}+...+\frac{1}{97}\right)-\left(\frac{1}{5}+\frac{1}{7}+...+\frac{1}{97}\right)\right]\)
\(\Rightarrow2S=\left(\frac{33}{99}-\frac{1}{99}\right)+0\)
\(\Rightarrow2S=\frac{32}{99}\)
\(\Rightarrow S=\frac{32}{99}\div2\)
\(\Rightarrow S=\frac{16}{99}\)
\(S=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{97.99}\)
\(S=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(S=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(S=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(S=\frac{1}{2}.\frac{32}{99}=\frac{16}{99}\)
2/3.5+ 2 /5.7+ 2/7.9+...+ 2/97.99
=1/3 - 1/5 + 1/5 - 1 /7 +.... + 1/97 - 1/99
=1/3 - 1/99
=32/99
m=/3.5+2/5.7+2/7.9+.....+2/97.99
=m=1/3-1/5+1/5-1/7+.......+1/97-1/99
m=1/3-1/99
=32/99
Mk bik câu B nè!
2B = 2/3.5 + 2/5.7 + 2/7.9 +.......+2/97.99
2B = 1/3 - 1/5 + 1/5 - 1/7 +.......+ 1/97 - 1/99
2B = 1/3 - 1/99
2B = 32/99
=> B = 16/99
=1/3-1/5+1/5-1/7+1/7-1/9+....+1/97-1/99
= 1/3 -1/99
=32/99
tích cho mình nha
=1/3-1/5+1/7-1/7+1/9-1/9+...+1/97-1/99
=1/3-1/99
=32/99
\(S=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{97\cdot99}\)
\(S=\frac{1}{2}\cdot\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{97\cdot99}\right)\)
\(S=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(S=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(S=\frac{1}{2}\cdot\frac{32}{99}\)
\(S=\frac{16}{99}\)
B = 1/3.5 + 1/.5.7 + 1/7.9 + 1/97.99
2B = 2/3.5 + 2/5.7 + 2.7.9 + 2/97.99
= 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/97 - 1/99
= 1/3 - 1/99
= 32 / 99
suy ra B = 32/99 : 2 = 16/99