tính
\(\left(\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}\right)\left(1+\frac{1}{2}+...+\frac{1}{2008}\right)\)
\(-\left(1+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}\right)\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}\right)\)
Những câu hỏi liên quan
Thu gọn
\(A=\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(2009^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(2010^4+\frac{1}{4}\right)}\)
\(B=\frac{\left(a+2008\right)!+\left(a+2009\right)!}{\left(a+2008\right)!-\left(a+2009!\right)}\)
Tính:
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{2008}-1\right)\left(\frac{1}{2009}-1\right)\)
Tính giá trị của biểu thức:Tfrac{4}{2.4}+frac{4}{4.6}+frac{4}{6.8}+...+frac{4}{2008.2010}Afrac{1}{1.3}+frac{1}{3.5}+frac{1}{5.7}+...+frac{1}{2007.2009}+frac{1}{2009.2011}Cleft(1-frac{1}{2}right)left(1-frac{1}{3}right)left(1-frac{1}{4}right)...left(1-frac{1}{1000}right)Sfrac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}Bleft(1+frac{1}{2}right)left(1+frac{1}{3}right)left(1+frac{1}{4}right)...left(1+frac{1}{100}right)Dleft(1-frac{1}{17}right)left(1-frac{2}{17}right)left(1-frac{3}{17}right)...left(1-f...
Đọc tiếp
Tính giá trị của biểu thức:
\(T=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009.2011}\)
\(C=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{1000}\right)\)
\(S=\frac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}\)
\(B=\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{100}\right)\)
\(D=\left(1-\frac{1}{17}\right)\left(1-\frac{2}{17}\right)\left(1-\frac{3}{17}\right)...\left(1-\frac{27}{17}\right)\)
\(T=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(T=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(T=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(T=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(T=2.\frac{502}{1005}=\frac{1004}{1005}\)
\(\Rightarrow T=\frac{1004}{1005}\)
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\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009+2011}\)
\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2009+2011}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)
\(A=\frac{1}{2}.\frac{2010}{2011}\)
\(\Rightarrow A=\frac{1005}{2011}\)
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\(C=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{100}\right)\)
\(C=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}\)
\(C=\frac{1.2.3...99}{2.3.4...100}\)
\(\Rightarrow C=\frac{1}{100}\)
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Xem thêm câu trả lời
Tính nhanh:
\(\frac{2009.\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2007}+\frac{1}{2008}\right)}{2008-\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{2006}{2007}+\frac{2007}{2008}\right)}\)
Thánh nào giải được thì làm ơn làm từng bước một nhé
Mong được chỉ giáo
THU GỌN\(A=\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(2009^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(2010^4+\frac{1}{4}\right)}\)
a, Tính : frac{left(13frac{1}{4}-2frac{5}{27}-10frac{5}{6}right).230frac{1}{25}+46frac{3}{4}}{left(1frac{3}{10}+frac{10}{3}right):left(12frac{1}{3}-14frac{2}{7}right)}b, Tính : Pfrac{frac{1}{2}+frac{1}{3}+frac{1}{4}+...+frac{1}{2012}}{frac{2011}{1}+frac{2010}{2}+frac{2009}{3}+...+frac{1}{2011}}c, Tính : frac{left(1+2+3+...+99+100right)left(frac{1}{2}-frac{1}{3}-frac{1}{7}-frac{1}{9}right)left(63.1,2-21.3,6right)}{1-2+3-4+...+99-100}
Đọc tiếp
a, Tính : \(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
b, Tính : \(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
c, Tính : \(\frac{\left(1+2+3+...+99+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)
Tính P = \(\left(1+\frac{1}{3}\right)\times\left(1+\frac{1}{2\times4}\right)\times\left(1+\frac{1}{3\times5}\right)\times\left(1+\frac{1}{4\times6}\right)\times...\times\left(1+\frac{1}{2009\times2011}\right)\)
(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
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Tính giá trị của biểu thức:
\(C=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{1000}\right)\)
\(S=\frac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}\)
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)
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Thực hiện phép tính :
a, \(S=2^{2010}-2^{2009}-2^{2008}-...-2-1\)
b, \(P=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\)
\(S=2^{2010}-\left(2^{2009}+2^{2008}+...+2+1\right)\)
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:
\(P=1+\dfrac{1+2}{2}+\dfrac{1+2+3}{3}+...+\dfrac{1+2+3+...+16}{16}\)
\(P=1+\dfrac{2.3}{2.2}+\dfrac{3.4}{2.3}+\dfrac{4.5}{2.4}+...+\dfrac{16.17}{2.16}\)
\(P=1+\dfrac{1}{2}\left(3+4+5+...+17\right)\)
\(P=1+\dfrac{1}{2}.\dfrac{\left(17-3+1\right)\left(3+17\right)}{2}=76\)
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