tính
\(\left(\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}\right)\left(1+\frac{1}{2}+...+\frac{1}{2008}\right)\)
\(-\left(1+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}\right)\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}\right)\)
K
Khách
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Những câu hỏi liên quan
S
14 tháng 4 2019
\(T=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(T=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(T=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(T=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(T=2.\frac{502}{1005}=\frac{1004}{1005}\)
\(\Rightarrow T=\frac{1004}{1005}\)
S
14 tháng 4 2019
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009+2011}\)
\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2009+2011}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)
\(A=\frac{1}{2}.\frac{2010}{2011}\)
\(\Rightarrow A=\frac{1005}{2011}\)
11 tháng 12 2016
(1+\(\frac{1}{3}\)) x (1+\(\frac{1}{2x4}\)) x(1+\(\frac{1}{3x5}\))x(1+\(\frac{1}{4x6}\)) x .....x (1+ \(\frac{1}{2009x2011}\))
= \(\frac{2}{1x3}\)x \(\frac{2}{2x4}\)x \(\frac{2}{3x5}\)x \(\frac{2}{4x6}\)x....x \(\frac{2}{2009x2011}\)
= ..................
đến đây tự làm nhé
14 tháng 4 2019
a) C=\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{1000}\right)\)
\(C=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{999}{1000}\)\(=\frac{1.2.3...999}{2.3.4...1000}=\frac{1.\left(2.3.4....999\right)}{\left(2.3.4....999\right).1000}\)\(=\frac{1}{1000}\)
b) Đặt: A=\(1+2+2^2+2^3+...+2^{2008}\)
\(\Leftrightarrow2A=2+2^2+2^3+....+2^{2008}+2^{2009}\)
\(\Leftrightarrow2A-A=2^{2009}-1\)
\(\Leftrightarrow A=2^{2009}-1\)
\(\Rightarrow S=\frac{2^{2009}-1}{1-2^{2009}}\)\(=\frac{2^{2009}-1}{-\left(2^{2009}-1\right)}=\frac{1}{-1}=-1\)
vậy: S=(-1)
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 1 2019
\(S=2^{2010}-\left(2^{2009}+2^{2008}+...+2+1\right)\)
Đặt \(A=2^{2009}+2^{2008}+...+2+1\)
\(\Rightarrow2A=2^{2010}+2^{2009}+...+2^2+2\)
\(\Rightarrow2A-2^{2010}+1=2^{2009}+2^{2008}+...+2+1\)
\(\Rightarrow2A-2^{2010}+1=A\)
\(\Rightarrow A=2^{2010}-1\)
\(\Rightarrow S=2^{2010}-A=2^{2010}-\left(2^{2010}-1\right)=1\)
b/ Ta có công thức \(1+2+3+...+n=\dfrac{n\left(n+1\right)}{2}\)
Do đó:
\(P=1+\dfrac{1+2}{2}+\dfrac{1+2+3}{3}+...+\dfrac{1+2+3+...+16}{16}\)
\(P=1+\dfrac{2.3}{2.2}+\dfrac{3.4}{2.3}+\dfrac{4.5}{2.4}+...+\dfrac{16.17}{2.16}\)
\(P=1+\dfrac{1}{2}\left(3+4+5+...+17\right)\)
\(P=1+\dfrac{1}{2}.\dfrac{\left(17-3+1\right)\left(3+17\right)}{2}=76\)
câu hỏi hay......nhưng tui xin nhường cho các bn khác
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