-1/3-1/15-...-1/33.35
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Những câu hỏi liên quan
NT
0
HH
1
AT
18 tháng 7 2017
a, Theo bài ra ta có:
\(M=\dfrac{2007}{1}+1+\dfrac{2006}{2}+1+.......+\dfrac{2}{2006}+1+\dfrac{1}{2007}+1-2007\)
( Ta thêm 1 vào mỗi một số hạng trong M nên phải bớt đi 2017 vì có 2017 số hạng ) ;'
\(=>M=2008+\dfrac{2008}{2}+\dfrac{2008}{3}+......+\dfrac{2008}{2007}+\dfrac{2008}{2007}-2007\)
\(=>M=\dfrac{2008}{2}+\dfrac{2008}{3}+\dfrac{2008}{4}+.....+\dfrac{2008}{2006}+\dfrac{2008}{2007}+1\)
Ta thấy xuất hiện 2008 chung nên đặt ra ngoài ta có:
\(=>M=2008\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+....+\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}\right)\)
\(=>M:N=2008\)
Câu b đợi 1 chút nha.......
18 tháng 7 2017
b, \(M=\dfrac{1}{11.13}+\dfrac{1}{13.15}+...+\dfrac{1}{31.33}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{11.13}+\dfrac{2}{13.15}+...+\dfrac{2}{31.33}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{31}-\dfrac{1}{33}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{11}-\dfrac{1}{33}\right)\)
\(=\dfrac{1}{33}\)
\(N=\dfrac{12}{11.13.15}+\dfrac{12}{13.15.17}+...+\dfrac{12}{31.33.35}\)
\(=3\left(\dfrac{4}{11.13.15}+\dfrac{4}{13.15.17}+...+\dfrac{4}{31.33.35}\right)\)
\(=3\left(\dfrac{1}{11.13}-\dfrac{1}{13.15}+\dfrac{1}{13.15}-\dfrac{1}{15.17}+...+\dfrac{1}{31.33}-\dfrac{1}{33.35}\right)\)
\(=3\left(\dfrac{1}{11.13}-\dfrac{1}{33.35}\right)\)
\(=\dfrac{92}{5005}\)
\(\Rightarrow M:N=\dfrac{1}{33}:\dfrac{92}{5005}=\dfrac{455}{276}\)
Vậy...
DL
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
2 tháng 8 2023
34 . 71 - 34 . 29
34 . ( 71 - 29 )
34 . -441
= -35721
(33 . 52 - 24 - 16 ) . 13
= ( 27 . 25- 16 - 16 ) . 13
= ( 675 - 16 -16 ) . 13
= 643. 13
= 8359
35.237 +33.35
=35 . ( \(237+33\) )
= 35 . 270
= 9450
23 . 42 + 23 . 84 - 40
=23 . ( 16 + 84 ) - 40
= 8 . 100 - 40
= 800 - 40
= 760
10 tháng 2 2019
Trả lời:
\(\frac{34}{34}>\frac{33}{35}\)
\(\frac{1991}{1999}< \frac{1995}{1995}\)
Học tốt
S
18 tháng 7 2017
a, Ta có:
\(M=\frac{1}{11.13}+\frac{1}{13.15}+...+\frac{1}{33.35}\)
\(=\frac{1}{2}\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{33}-\frac{1}{35}\right)\)
\(=\frac{1}{2}\left(\frac{1}{11}-\frac{1}{35}\right)=\frac{1}{2}\cdot\frac{24}{385}=\frac{12}{385}\)
\(N=\frac{12}{11.13.15}+\frac{12}{13.15.17}+...+\frac{12}{31.33.35}\)
\(=3\left(\frac{1}{11.13}-\frac{1}{13.15}+\frac{1}{13.15}-\frac{1}{15.17}+...+\frac{1}{31.33}-\frac{1}{33.35}\right)\)
\(=3\left(\frac{1}{11.13}-\frac{1}{33.35}\right)=3\cdot\frac{92}{15015}=\frac{92}{5005}\)
\(\Rightarrow M:N=\frac{12}{385}:\frac{92}{5005}=\frac{39}{23}\)
b, \(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}-\left(1+\frac{1}{2}+...+\frac{1}{1006}\right)\)
\(=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2013}\)
\(\Rightarrow S-P-1=\left(\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2013}\right)-\left(\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2013}\right)-1=0-1=-1\)
Q
0
\(=\dfrac{1}{3}-\dfrac{1}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{33\cdot35}\right)\)
\(=\dfrac{1}{3}-\dfrac{1}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{33}-\dfrac{1}{35}\right)\)
\(=\dfrac{1}{3}-\dfrac{1}{2}\cdot\dfrac{32}{105}=\dfrac{19}{105}\)