LT
1
K
Khách
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NV
0
NM
3
21 tháng 3 2017
E=\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{98.99.100}\)
* Áp dụng công thức: \(\dfrac{k}{n.\left(n+k\right)}\)=\(\dfrac{1}{n}-\dfrac{1}{n+k}\)
ta có : \(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-....+\dfrac{1}{98.99}-\dfrac{1}{99.100}\)
E=\(\dfrac{1}{1.2}-\dfrac{1}{99.100}\)
E= ........(tính ra)
KH
12 tháng 5 2017
* Chứng tỏ
Ta có :\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\)
= \(\dfrac{1}{1.2.3}.\dfrac{2}{2}+\dfrac{1}{2.3.4}.\dfrac{2}{2}+...+\dfrac{1}{98.99.100}.\dfrac{2}{2}\)
= \(\dfrac{1}{2}.\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{98.99.100}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}+0+0+...+0+\dfrac{-1}{99.100}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{2}+\dfrac{-1}{9900}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{4850}{9900}+\dfrac{-1}{9900}\right)\)
= \(\dfrac{1}{2}.\dfrac{4849}{9900}\)
= \(\dfrac{4849}{19800}\)
4 tháng 3 2017
a) Ta có: \(3xy+x-3y=6\)
\(\Rightarrow x\left(3y+1\right)-3y=6\)
\(\Rightarrow x\left(3y+1\right)-\left(3y+1\right)=5\)
\(\Rightarrow\left(x-1\right)\left(3y+1\right)=5\)
Ta có bảng sau:
....
b) Ta có: \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(=\frac{1}{2}.\frac{4949}{9900}\)
\(=\frac{4949}{19800}\)
\(\Rightarrow\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}=\frac{4949}{19800}\left(đpcm\right)\)
Vậy...
NL
1
17 tháng 5 2022
\(2S=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{23+24+25}=\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}\right)+\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)+...+\left(\dfrac{1}{23.24}-\dfrac{1}{24.25}\right)\)\(=\dfrac{1}{1.2}-\dfrac{1}{24.25}=\dfrac{299}{600}\)
Vậy \(S=\dfrac{299}{600}\div2=\dfrac{299}{1200}\)
TQ
1
QT
Quoc Tran Anh Le
Giáo viên
3 tháng 12 2017
\(S=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{8.9.10}\)
\(S=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{8.9}-\dfrac{1}{9.10}\)
\(S=\dfrac{1}{1.2}-\dfrac{1}{9.10}\)
\(S=\dfrac{1}{2}-\dfrac{1}{90}=\dfrac{44}{90}\)
22 tháng 4 2021
Tìm y:
-y:1/2-5/2=4+1/2
-y:1/2 = 4+1/2+5/2
-y:1/2 = 7
-y = 7.2
y = -14
Vậy y = -14
1 tháng 6 2017
S = \(\dfrac{1}{1.4}\)+ \(\dfrac{1}{4.7}\)+...+\(\dfrac{1}{2002.2005}\)
S = ( 1 - \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-...+\(\dfrac{1}{2002}\)-\(\dfrac{1}{2005}\)) . \(\dfrac{1}{3}\)
S = ( 1 - \(\dfrac{1}{2005}\)) . \(\dfrac{1}{3}\)
S = \(\dfrac{2004}{2005}\). \(\dfrac{1}{3}\)
S = \(\dfrac{2014}{6015}\)
1 tháng 6 2017
a) \(S=\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{2002.2005}\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2002}-\dfrac{1}{2005}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{2005}\right)\)
\(=\dfrac{1}{3}.\dfrac{2004}{2005}=\dfrac{668}{2005}\)
KL.
b) \(P=\dfrac{3}{1.6}+\dfrac{3}{6.11}+\dfrac{3}{11.16}+...+\dfrac{3}{96.101}\)
\(=\dfrac{3}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{96}-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{5}\left(1-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{5}.\dfrac{100}{101}=\dfrac{60}{101}\)
KL.
c) \(Q=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{99.100}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{9900}\right)\)
\(=\dfrac{1}{2}.\dfrac{1}{19800}=\dfrac{1}{39600}\)
KL.
GL
2
YN
2 tháng 2 2023
\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2004.2005.2006}\)
\(=2.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}\right)+2.\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)+...+2.\left(\dfrac{1}{2004.2005}-\dfrac{1}{2005.2006}\right)\)
\(=2.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{2004.2005}-\dfrac{1}{2005.2006}\right)\)
\(=2.\left(\dfrac{1}{1.2}-\dfrac{1}{2005.2006}\right)\)
\(=1-\dfrac{2}{2005.2006}\)
\(=\dfrac{2011014}{2011015}\).
2 tháng 2 2023
Ta có:
\(M=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2004.2005.2006}\)
\(M=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{2004.2005.2006}\right)\)
\(M=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2004.2005}-\dfrac{1}{2005.2006}\right)\)
\(M=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2005.2006}\right)\)
Ta có :
\(S=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+..............+\dfrac{1}{98.99.100}\)
\(S=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+................+\dfrac{2}{98.99.100}\right)\)
\(S=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...........+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)\)
\(S=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{99.100}\right)\)
\(S=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{9900}\right)\)
\(S=\dfrac{1}{2}.\dfrac{4949}{9900}\)
\(S=\dfrac{4949}{19800}\)
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