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\(2005a=\frac{2005^{2006}+2005}{2005^{2006}+1}=\frac{2005^{2006}+1}{2005^{2006}+1}+\frac{2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}\)
\(2005b=\frac{2005^{2005}+2005}{2005^{2005}+1}=\frac{2005^{2005}+1}{2005^{2005}+1}+\frac{2004}{2005^{2005}+1}=1+\frac{2004}{2005^{2005}+1}\)
Ta thấy :\(2005^{2006}+1>2005^{2005}+1\)
\(\Rightarrow\frac{2004}{2005^{2006}+1}< \frac{2004}{2005^{2005}+1}\)
\(\Rightarrow1+\frac{2004}{2005^{2006}+1}< 1+\frac{2004}{2005^{2005}+1}\)
\(\Rightarrow2005a< 2005b\)
\(\Rightarrow a< b\)
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N=\(\frac{-7}{10^{2005}}+\frac{-15}{10^{2005}}\) Và M=\(\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}\)
Ta xét 2 PS \(\frac{-7}{10^{2005}}\) và \(\frac{-7}{10^{2006}}\)
Ta có tích . (-7).102006<(-7).102005 (vì 102006>102005)
Nên \(\frac{-7}{10^{2005}}\) < \(\frac{-7}{10^{2006}}\)
Nên \(\frac{-7}{10^{2005}}+\frac{-15}{10^{2005}}\) < \(\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}\)
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ta cs: \(\frac{a+2006}{a-2006}=\frac{b+2005}{b-2005}\)
\(\Rightarrow\frac{a+2006}{b+2005}=\frac{a-2006}{b-2005}=\frac{a}{b}=\frac{2006}{2005}\)
=> dpcm
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1 + 22 + 23 + ... + 22005
Gọi dãy số trên là A
A = \(1+2^2+2^3+....+2^{2005}\)
A =\(2^0+2^2+2^3+....+2^{2005}\)
A + \(2^1\)= \(2^0+2^1+2^2+2^3+....+2^{2005}\)
( A + 2 ) x 21 = \(\left(2^0+2^1+2^2+2^3+....+2^{2005}\right)\times2^1\)
Ax2 + 4 =\(2^1+2^2+2^3+2^4+....+2^{2006}\)
4 + A x 2 - A =\(2^1+2^2+2^3+2^4+....+2^{2006}-\left(1+2^2+2^3+...2^{2005}\right)\)
4 + A = \(2^1+2^2+2^3+2^4+....+2^{2006}-1-2^2-2^3-....-2^{2005}\)
4 + A = \(2^{2006}-1\)
A=\(2^{2006}-1-4\)
A = \(2^{2006}-5\)
Mà \(2^{2006}-5< 2^{2006}\)
\(\Rightarrow1+2^2+2^3+....+2^{2005}< 2^{2006}\)
\(\sqrt{2005+2006}^2=2005+2006=4011\)
\(\left(\sqrt{2005}+\sqrt{2006}\right)^2=2005+2\sqrt{2005}.\sqrt{2006}+2006=4011+2\sqrt{2005}.\sqrt{2006}\)
Vì \(2\sqrt{2005}.\sqrt{2006}>0\) nên =>\(4011
ai biết thì giải giúp với