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30 tháng 4 2016

200 nha

30 tháng 4 2016

200 nha 

nhé

9 tháng 4 2018

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{49.50.51}\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{49.50}-\frac{1}{50.51}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{50.51}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2550}\right)\)

\(=\frac{1}{2}.\left(\frac{1275}{2550}-\frac{1}{2550}\right)\)

\(=\frac{1}{2}.\frac{1274}{2550}\)

\(=\frac{637}{2550}\)

Lưu ý : Dấu \("."\)là dấu \("\)x  \("\)

( dấu nhân ) 

Chúc bạn học giỏi !!! 

9 tháng 4 2018

Công thức : 

\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{1}{2}.\frac{2}{6}=\frac{1}{6}=\frac{1}{1.2.3}\)

VD ( dễ hiểu ) 

24 tháng 3 2022

Ta có:

\(A=\frac{1}{1\text{x}2\text{x}3}+\frac{1}{2\text{x}3\text{x}4}+\frac{1}{3\text{x}4\text{x}5}+...+\frac{1}{18\text{x}19\text{x}20}< \frac{1}{4}\)

\(A=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{20}< \frac{1}{4}\)

\(A=1+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+...+\frac{1}{20}< \frac{1}{4}\)

\(A=1+\frac{1}{20}< \frac{1}{4}\)

\(A=\frac{19}{20}< \frac{1}{4}\)

\(A=\frac{19}{20}< \frac{5}{20}\)

\(A>\frac{1}{4}\)

20 tháng 10 2024

Cyak 3ampo

17 tháng 8 2018

A =1x2x3 + 2x3x4 +3x4x5+....+ 2010 x2011 x 2012

4A =1x2x3x4 + 2x3x4x4 +3x4x5x4+....+ 2010 x2011 x 2012x4

4A =1x2x3x4 + 2x3x4x(5+1) +3x4x5x(6-2)+....+ 2010 x2011 x 2012x(2013-2009)

4A =1x2x3x4 + 2x3x4x5-1x2x3x4+3x4x5x6-2x3x4x5+....+ 2010 x2011 x 2012x2013-2009x2010x2011x2012

4A = 2010 x2011 x 2012x2013

A = \(\frac{2010\times2011\times2012\times2013}{4}\)

17 tháng 8 2018

đặt S=1.2.3+2.3.4+....+18.19.20

4S=1.2.3.4+2.3.4.(5-1)+.......+18.19.20.(21-17)

4S=1.2.3.4-1.2.3+2.3.4.5-1.2.3.4+......+18.19.20.21-17.18.19.20

4S=....tự làm nha

17 tháng 11 2018

Đặt \(A=\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{30\times31\times32}\)

\(2A=\frac{2}{1\times2\times3}+\frac{2}{2\times3\times4}+\frac{2}{3\times4\times5}+...+\frac{2}{30\times31\times32}\)

\(=\left(\frac{1}{1\times2}-\frac{1}{2\times3}\right)+\left(\frac{1}{2\times3}-\frac{1}{3\times4}\right)+\left(\frac{1}{3\times4}-\frac{1}{4\times5}\right)+...+\left(\frac{1}{30\times31}-\frac{1}{31\times32}\right)\)

\(=\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+\frac{1}{3\times4}-\frac{1}{4\times5}+...+\frac{1}{30\times31}-\frac{1}{31\times32}\)

\(=\frac{1}{1\times2}-\frac{1}{31\times32}\)

\(=\frac{1}{2}-\frac{1}{992}\)

\(=\frac{495}{992}\)

\(\Rightarrow A=\frac{495}{992}\div2=\frac{495}{1984}\)

17 tháng 11 2018

\(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{30\times31\times32}\)

\(=\frac{1}{2}\times\left(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{30\times31\times32}\right)\)

\(=\frac{1}{2}\times\left(\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+\frac{1}{3\times4}-\frac{1}{4\times5}+...+\frac{1}{30\times31}-\frac{1}{31\times32}\right)\)

\(=\frac{1}{2}\times\left(\frac{1}{1\times2}-\frac{1}{31\times32}\right)\)

\(=\frac{1}{2}\times\frac{990}{1984}\)

\(=\frac{990}{3968}=\frac{495}{1984}\)

21 tháng 9 2015

S = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2013.2014.2015}\)

S = \(\frac{1}{2}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+....+\frac{2015-2013}{2013.2014.2015}\right)\)

S = \(\frac{1}{2}.\left(\frac{3}{1.2.3}-\frac{1}{1.2.3}+\frac{4}{2.3.4}-\frac{2}{2.3.4}+...+\frac{2015}{2013.2014.2015}-\frac{2013}{2013.2014.2015}\right)\)

S = \(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2013.2014}-\frac{1}{2014.2015}\right)\)

S = \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2014.2015}\right)\)

S = \(\frac{1}{2}.\frac{2029104}{4058210}\)

S = \(\frac{1014552}{4058210}\)

3 tháng 8 2016

gcjjjjjjjjjjjhm.f

17 tháng 11 2021

Vũ Thị Trang lại là một nạn nhân tiếp tục bị báo cáo

10 tháng 4 2024

A = \(\dfrac{2}{1\times3\times5}\) + \(\dfrac{2}{3\times5\times7}\) + \(\dfrac{2}{5\times7\times9}\)+\(\dfrac{2}{7\times9\times11}\)

A = \(\dfrac{1}{2}\) x (\(\dfrac{4}{1\times3\times5}\) + \(\dfrac{4}{3\times5\times7}\) + \(\dfrac{4}{5\times7\times9}\) + \(\dfrac{4}{7\times9\times11}\))

A = \(\dfrac{1}{2}\)x (\(\dfrac{1}{1\times3}\)-\(\dfrac{1}{3\times5}\)+\(\dfrac{1}{3\times5}\)-\(\dfrac{1}{5\times7}\)+\(\dfrac{1}{5\times7}\)-\(\dfrac{1}{7\times9}\)+\(\dfrac{1}{7\times9}\)-\(\dfrac{1}{9\times11}\))

A = \(\dfrac{1}{2}\)x (\(\dfrac{1}{1\times3}\) - \(\dfrac{1}{9\times11}\))

A = \(\dfrac{1}{2}\) x (\(\dfrac{1}{3}-\dfrac{1}{99}\))

A = \(\dfrac{1}{2}\times\) \(\dfrac{32}{99}\)

A = \(\dfrac{16}{99}\)

10 tháng 4 2024

B = \(\dfrac{1}{1\times2\times3}\) + \(\dfrac{1}{2\times3\times4}\) + \(\dfrac{1}{3\times4\times5}\) + \(\dfrac{1}{4\times5\times6}\)

B = \(\dfrac{1}{2}\) x (\(\dfrac{2}{1\times2\times3}+\dfrac{2}{2\times3\times4}+\dfrac{2}{3\times4\times5}+\dfrac{2}{4\times5\times6}\))

B = \(\dfrac{1}{2}\) x (\(\dfrac{1}{1\times2}\)-\(\dfrac{1}{2\times3}\) + \(\dfrac{1}{2\times3}\)-\(\dfrac{1}{3\times4}\)+\(\dfrac{1}{3\times4}\)-\(\dfrac{1}{4\times5}\)+\(\dfrac{1}{4\times5}\)-\(\dfrac{1}{5\times6}\))

B = \(\dfrac{1}{2}\)x(\(\dfrac{1}{1\times2}\) - \(\dfrac{1}{5\times6}\))

B = \(\dfrac{1}{2}\)x (\(\dfrac{1}{2}-\dfrac{1}{30}\))

B = \(\dfrac{1}{2}\)\(\dfrac{7}{15}\)

B = \(\dfrac{7}{30}\)