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= 1/3 x 5 + 1/5x 7 + 1/7 x 9 +...+1/99 x 101
=1/ 2x (1/3 - 1/5 +1/5 - 1/7 +1/7 - 1/9 + 1/99 - 1/101)
=1/2 x (1/3 - 1/99)
=1/2 x (1/3 - 1/101)
=1/2 x (98/303)
=1/15 + 1/35 + 1/63 +1/99+...+1/9999
=49/303
\(=\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{99.101}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{3}-\frac{1}{101}+0+...+0\)
\(=\frac{98}{303}\)
\(\Rightarrow A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{99.101}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{99.101}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{88}{303}\)
\(\Rightarrow A=\frac{44}{303}\)
\(A=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{99\times101}\)
\(\Rightarrow2A=\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+...+\frac{2}{99\times101}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\)
=> A = 98/203 : 2 = 49/303
=> 2(1/15+1/35+1/63+1/99)x=2
=>(1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11)x=2
=>8/33x=2
=>x=2:8/33
=>x=8,25
\(\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\cdot x=1\)
\(\left(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}\right)\cdot x=1\)
\(\left[\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\right]\cdot x=1\)
\(\left[\frac{1}{2}\left(\frac{1}{3}-\frac{1}{11}\right)\right]\cdot x=1\)
\(\left[\frac{1}{2}\cdot\frac{8}{33}\right]\cdot x=1\)
\(\frac{4}{33}\cdot x=1\)
\(\Rightarrow x=\frac{1}{\frac{4}{33}}=\frac{33}{4}\)
\(2B=\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{9.11}\)
\(2B=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\)
\(2B=\frac{1}{1}-\frac{1}{11}=\frac{10}{11}\)
\(B=\frac{10}{11}:2=\frac{10}{11}.\frac{1}{2}=\frac{5}{11}\)
\(B=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{1}{2}.\frac{10}{11}=\frac{5}{11}\)
<=> \(\left(\frac{1}{3\cdot5}+\frac{1}{5.7}+...+\frac{1}{13\cdot15}\right)+x=\frac{17}{15}\)
<=> \(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...-\frac{1}{15}\right)+x=\frac{17}{15}\)
<=>\(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{15}\right)+x=\frac{17}{15}\)
<=> \(\frac{2}{15}+x=\frac{17}{15}\)
=> x = 1
(1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15)+x=17/15
[2.(1/3-1/5+1/5-1/7+...+1/13-1/15)]+x=17/15
[2.(1/3-1/15)]+x=17/15
(2.4/15)+x=17/15
6/15+x=17/15
x=17/15-6/15
x=11/15
Đăt S=1/15+1/35+1/63+1/99+...+1/2915+1/3135
=1/3.5+1/5.7+1/7.9+1/9.11+...+1/53.55+1/55.57
=1/2(2/3.5+2/5.7+2/7.9+...+2/53.55+2/55.57)
=1/2(1/3-1/5+1/5-1/7+1/7-1/9+...+1/53-1/55+1/55-1/57)
=1/2(1/3-1/57)
=1/2(19/57-1/57)
=1/2.18/57
=3/19
Vậy 1/15+1/35+1/63+1/99+...+1/2915+1/3135=3/19
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Đặt \(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{2915}+\frac{1}{3135}\)
\(\Leftrightarrow A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+....+\frac{1}{53\cdot55}+\frac{1}{55\cdot57}\)
\(\Leftrightarrow2A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+...+\frac{2}{53\cdot55}+\frac{2}{55\cdot57}\)
\(\Leftrightarrow2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-....+\frac{1}{53}-\frac{1}{55}+\frac{1}{55}-\frac{1}{57}\)
\(\Leftrightarrow2A=\frac{1}{3}-\frac{1}{57}=\frac{6}{19}\)
\(\Leftrightarrow A=\frac{6}{19}:2=\frac{3}{19}\)
Dấu \(.\)là dấu nhân
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\frac{1}{2}.\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}+\frac{2}{195}\right)\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{15}\right)\)
\(=\frac{1}{2}.\frac{14}{15}\)
\(=\frac{7}{15}\)
~ Ủng hộ nhé
Đặt \(A=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)
Suy ra ; \(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{13}-\frac{1}{15}\)
\(=1-\frac{1}{15}=\frac{14}{15}\)
=> A = \(\frac{14}{15}:2=\frac{14}{15}.\frac{1}{2}=\frac{7}{15}\)
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