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Đặt biểu thực là A
2A = \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{21.23}\)
= \(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{21}-\dfrac{1}{23}=1-\dfrac{1}{23}=\dfrac{22}{23}\)
=> A = \(\dfrac{11}{23}\)
Đề sai rồi em, mẫu số đều là số lẻ thì 120 ko theo quy luật
\(A=\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+...+\dfrac{1}{899}\\ 2A=2\cdot\dfrac{1}{3}+2\cdot\dfrac{1}{15}+2\cdot\dfrac{1}{35}+2\cdot\dfrac{1}{63}+...+2\cdot\dfrac{1}{899}\\ 2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{29}-\dfrac{1}{31}\\ 2A=1-\dfrac{1}{31}\\ 2A=\dfrac{30}{31}\\ A=\dfrac{30}{31}\div2\\ A=\dfrac{30}{31\cdot2}=\dfrac{15}{31}\)
:))
\(\Leftrightarrow\dfrac{4}{9}:x=\dfrac{8}{3}\)
hay \(x=\dfrac{4}{9}\cdot\dfrac{3}{8}=\dfrac{1}{2}\cdot\dfrac{1}{3}=\dfrac{1}{6}\)
\(C=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{1023}\)
\(=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}....+\frac{1}{31\cdot33}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{31}-\frac{1}{33}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{33}\right)\)
\(=\frac{1}{2}\cdot\frac{32}{33}\)
\(=\frac{32}{66}=\frac{16}{33}\)
Vậy \(A=\frac{16}{33}\)
HOK TỐT .
\(C=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{1023}\)
\(C=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{31\cdot33}\)
\(C=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{31\cdot33}\right)\)
\(C=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{31}-\frac{1}{33}\right)\)
\(C=\frac{1}{2}\left(1-\frac{1}{33}\right)\)
\(C=\frac{1}{2}\cdot\frac{32}{33}\)
\(C=\frac{16}{33}\)
1/3+1/15+1/35+1/63+1/99+……+1/9999
=1/(1×3)+1/(3×5)+1/(5×7)+1/(7×9)+1/(9×11)+……+1/(99×101)
=1/2(1-1/3)+1/2(1/3-1/5)+1/2(1/5-1/7)+1/2(1/7-1/9)+1/2(1/9-1/11)+……+1/2(1/99-1/101)
=1/2(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+……+1/99-1/101)
=1/2(1-1/101)
=1/2×(100/101)
=50/101
1/3+1/15+1/35+1/63+1/99+……+1/9999
=1/(1×3)+1/(3×5)+1/(5×7)+1/(7×9)+1/(9×11)+……+1/(99×101)
=1/2(1-1/3)+1/2(1/3-1/5)+1/2(1/5-1/7)+1/2(1/7-1/9)+1/2(1/9-1/11)+……+1/2(1/99-1/101)
=1/2(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+……+1/99-1/101)
=1/2(1-1/101)
=1/2×(100/101)
=50/101
1/1x3 + 1/3x5 + 1/5x7 + 1/7x9 +... + 1/23x25 + 1/25x27 =
1/2 x (2/1x3 + 2/3x5 + 2/5x7 +..... 2/23x25 +2/25x27 =
1/2 x (1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 +...... + 1/23 - 1/25 + 1/25 - 1/27 =
1/2 x (1 - 1/27) =
1/2 x 26/27 = 13/27
\(A=\frac{1}{1.3}+\frac{1}{3.5}+....+\frac{1}{21.23}\)
A=\(\frac{1}{3}\left(1-\frac{1}{3}+\frac{1}{3}-.......-\frac{1}{23}\right)\)=\(\frac{1}{3}.\frac{22}{23}=\frac{22}{69}\)
hok t
tl lại
\(A=\frac{1}{1.3}+....\frac{1}{21.23}\)
\(A=\frac{1}{2}\left(1-\frac{1}{3}+.....+\frac{1}{21}-\frac{1}{23}\right)\)
\(A=\frac{1}{2}.\frac{22}{23}=\frac{11}{23}\)
k t nha