a,A=1-2+3+4-5-6+7+8-9-...+2007+2008-2009-2010
b, \(\frac{1}{5^2}-\frac{1}{5^3}+\frac{1}{5^4}-\frac{1}{5^5}+..-\frac{1}{5^{101}}\).CM<\(\frac{1}{30}\)
Những câu hỏi liên quan
Tính:
Afrac{7}{3}.frac{11}{16}+frac{10}{3}.frac{7}{16}-frac{7}{6}.frac{5}{8}
B1+2-3-4+5+6-7-8+.....+2005+2006-2007-2008+2009+2010
C(1-frac{1}{4})(1-frac{1}{9})(1-frac{1}{16})......(1-frac{1}{100000})
Dfrac{17frac{3}{4}.frac{17}{5}+3frac{2}{5}.82frac{1}{4}}{2.34-3.17}
Efrac{frac{2008}{2011}+frac{2009}{2010}+frac{2010}{2009}+frac{2011}{2008}+frac{2012}{503}}{frac{1}{2011}+frac{1}{2010}+frac{1}{2009}+frac{1}{2008}}
F(2-frac{2}{1.3})+(2-frac{2}{3.5})+(2-frac{2}{5.7})+.....+(2-frac{2}{2009.201...
Đọc tiếp
Tính:
A=\(\frac{7}{3}\).\(\frac{11}{16}\)+\(\frac{10}{3}\).\(\frac{7}{16}\)-\(\frac{7}{6}\).\(\frac{5}{8}\)
B=1+2-3-4+5+6-7-8+.....+2005+2006-2007-2008+2009+2010
C=(1-\(\frac{1}{4}\))(1-\(\frac{1}{9}\))(1-\(\frac{1}{16}\))......(1-\(\frac{1}{100000}\))
D=\(\frac{17\frac{3}{4}.\frac{17}{5}+3\frac{2}{5}.82\frac{1}{4}}{2.34-3.17}\)
E=\(\frac{\frac{2008}{2011}+\frac{2009}{2010}+\frac{2010}{2009}+\frac{2011}{2008}+\frac{2012}{503}}{\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}}\)
F=(2-\(\frac{2}{1.3}\))+(2-\(\frac{2}{3.5}\))+(2-\(\frac{2}{5.7}\))+.....+(2-\(\frac{2}{2009.2011}\))
Tính giá trị biểu thức
A=1-2+3+4-5-6+7+8-9-.....+2007+2008-2009-2010
B=\(1-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-...-\frac{1}{9900}\)
b: \(B=1-\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)\)
\(=1-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=1-\left(\dfrac{1}{2}-\dfrac{1}{100}\right)=\dfrac{1}{2}-\dfrac{49}{100}=\dfrac{1}{100}\)
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tính tổng sau :\(c=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{\frac{5}{2008}-\frac{5}{2009}-\frac{5}{2010}}+\)\(\frac{\frac{2}{2007}-\frac{2}{2008}-\frac{2}{2009}}{\frac{3}{2007}-\frac{3}{2008}-\frac{3}{2009}}\)
\(C=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{\frac{5}{2008}-\frac{5}{2009}-\frac{5}{2010}}+\frac{\frac{2}{2007}-\frac{2}{2008}-\frac{2}{2009}}{\frac{3}{2007}-\frac{3}{2008}-\frac{3}{2009}}\)
\(=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{5.\left(\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)}+\frac{2.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}{3.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}\)
\(=\frac{1}{5}+\frac{2}{3}\)
\(=\frac{13}{15}\)
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1.Tính tổngSleft(frac{-1}{7}right)^0+left(frac{-1}{7}right)^1+left(frac{-1}{7}right)^2+...+left(frac{-1}{7}right)^{2007}2.Tìm x5^x+5^{x+2}6503.CMRfrac{1}{6} frac{1}{5^2}+frac{1}{6^2}+frac{1}{7^2}+...+frac{1}{100^2} frac{1}{4}4. Cho Afrac{1}{2010}+frac{2}{2009}+frac{3}{2008}+...+frac{2009}{2}+frac{2010}{1} Bfrac{1}{2}+frac{1}{3}+frac{1}{4}+...+frac{1}{2010}+frac{1}{2011}So sánh A và B
Đọc tiếp
1.Tính tổng
\(S=\left(\frac{-1}{7}\right)^0+\left(\frac{-1}{7}\right)^1+\left(\frac{-1}{7}\right)^2+...+\left(\frac{-1}{7}\right)^{2007}\)
2.Tìm x
\(5^x+5^{x+2}=650\)
3.CMR
\(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\)
4. Cho \(A=\frac{1}{2010}+\frac{2}{2009}+\frac{3}{2008}+...+\frac{2009}{2}+\frac{2010}{1}\)
\(B=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2010}+\frac{1}{2011}\)
So sánh A và B
So sánh
A=\(\frac{2006}{2007}-\frac{2007}{2008}+\frac{2008}{2009}-\frac{2009}{2010}\)
B=\(-\frac{1}{2006.2007}-\frac{1}{2008.2009}\)
So sánh
B=\(-\frac{1}{2011}-\frac{7}{11^2}-\frac{5}{11^3}-\frac{3}{11^4}\)
A=\(-\frac{1}{2011}-\frac{3}{11^2}-\frac{5}{11^4}\)
7.Giá trị biểu thức A=\(\frac{2008+\frac{2007}{2}+\frac{2006}{3}+\frac{2005}{4}+...+\frac{2}{2007}+\frac{1}{2008}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}\) là A=.................
\(A=\frac{2008+\frac{2007}{2}+\frac{2006}{3}+\frac{2005}{4}+...+\frac{2}{2007}+\frac{1}{2008}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}\)
Gọi a là tử số, b là mẫu số của phân số A
a = \(\frac{2008}{1}\)+ \(\frac{2007}{2}\)+ \(\frac{2006}{3}\)+ ... + \(\frac{1}{2008}\)
Dãy số a có (2008 - 1) : 1 + 1 = 2008 số. Và a = ( \(\frac{2008}{1}\)+ \(\frac{1}{2008}\)) x (2008 : 2)
b = \(\frac{1}{2}\)+ \(\frac{1}{3}\)+ \(\frac{1}{4}\)+ ... + \(\frac{1}{2009}\)
Dãy số b có (2009 - 2) : 1 + 1 = 2008 số. Và b = (\(\frac{1}{2}\)+ \(\frac{1}{2009}\)) x (2008 : 2)
A = [ ( \(\frac{2008}{1}\)+ \(\frac{1}{2008}\)) x (2008 : 2)] : [ (\(\frac{1}{2}\)+ \(\frac{1}{2009}\)) x (2008 : 2)] = ( \(\frac{2008}{1}\)+ \(\frac{1}{2008}\)) : (\(\frac{1}{2}\)+ \(\frac{1}{2009}\))
A = \(\frac{\text{2008 x2008 + 1}}{2008}\)x \(\frac{2x2009+2}{2x2009}\)
A = 2008
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So sánh A và B biết:a.A-frac{1}{2016}-frac{3}{^{11^2}}-frac{5}{11^3}-frac{7}{11^4}B-frac{7}{11^2}-frac{1}{2016}-frac{5}{11^3}-frac{3}{11^4}b.Afrac{2006}{2007}-frac{2007}{2008}+frac{2008}{2009}-frac{2009}{2010}B-frac{1}{2006.2007}-frac{1}{2008.2009}Nhờ các bạn trình bày hộ mình mình sẽ tick cho. Cảm ơn
Đọc tiếp
So sánh A và B biết:
\(a.A=\)\(-\frac{1}{2016}-\frac{3}{^{11^2}}-\frac{5}{11^3}-\frac{7}{11^4}\)
\(B=-\frac{7}{11^2}-\frac{1}{2016}-\frac{5}{11^3}-\frac{3}{11^4}\)
\(b.A=\frac{2006}{2007}-\frac{2007}{2008}+\frac{2008}{2009}-\frac{2009}{2010}\)
\(B=-\frac{1}{2006.2007}-\frac{1}{2008.2009}\)
Nhờ các bạn trình bày hộ mình mình sẽ tick cho. Cảm ơn
\(A=\frac{2008+\frac{2007}{2}+\frac{2006}{3}+\frac{2005}{4}+...+\frac{2}{2007}+\frac{1}{2008}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}\)
\(A=\frac{2008+\frac{2007}{2}+\frac{2006}{3}+\frac{2005}{4}+...+\frac{2}{2007}+\frac{1}{2008}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}\)
\(=\frac{\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+\left(1+\frac{2005}{4}\right)+...+\left(1+\frac{1}{2007}\right)+\left(1+\frac{1}{2008}\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}\)
\(=\frac{\frac{2009}{2}+\frac{2009}{3}+\frac{2009}{4}+...+\frac{2009}{2007}+\frac{2009}{2008}+\frac{2009}{2009}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}\)
\(=\frac{2009\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}=2009\)
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