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\(B=4+3^2+3^3+...+3^{2004}\)
\(\Rightarrow B=1+3+3^2+3^3+...+3^{2004}\)
\(\Rightarrow3B=3+3^2+3^3+...+3^{2005}\)
\(\Rightarrow3B-B=3+3^2+3^3+...+3^{2005}-1-3-3^2-...-3^{2004}\)
\(\Rightarrow2B=3^{2005}-1\)
\(\Rightarrow B=\frac{3^{2005}-1}{2}< \frac{3^{2005}}{2}< 3^{2005}=C\)
Vậy B < C
\(A=\frac{2005^{2005}+1}{2005^{2006}+1}\)
\(2005A=\frac{2005^{2006}+2005}{2005^{2006}+1}=\frac{2005^{2006}+1+2004}{2005^{2006}+1}=\frac{2005^{2006}+1}{2005^{2006}+1}+\frac{2004}{2005^{2006}+1}\)
\(B=\frac{2005^{2004}+1}{2005^{2005}+1}\)
\(2005B=\frac{2005^{2005}+2005}{2005^{2005}+1}=\frac{2005^{2005}+1+2004}{2005^{2005}+1}=\frac{2005^{2005}+1}{2005^{2005}+1}+\frac{2004}{2005^{2005}+1}\)
Vì \(\frac{2004}{2005^{2006}+1}<\frac{2004}{2005^{2005}+1}\)
Nên A<B
\(N=\frac{2004+2005}{2005+2006}=\frac{2004}{2005+2006}+\frac{2005}{2005+2006}\)
\(\text{Vì }\frac{2004}{2005}>\frac{2004}{2005+2006};\frac{2005}{2006}>\frac{2005}{2005+2006}\text{nên:}\)
\(\frac{2004}{2005}+\frac{2005}{2006}>\frac{2004}{2005+2006}+\frac{2005}{2005+2006}\)
Vậy M>N
a) Có \(3^{125}=3^{124}.3=\left(3^4\right)^{31}.3=81^{31}.3\)
\(4^{93}=\left(4^3\right)^{31}=64^{31}\)
Vì \(81^{31}>64^{31}\Rightarrow81^{31}.3>64^{31}\)
=) \(3^{125}>4^{93}\)
b) Có \(A=\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}=\frac{-7}{10^{2005}}+\frac{-7}{10^{2006}}+\frac{-8}{10^{2006}}\)
\(B=\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}=\frac{-7}{10^{2005}}+\frac{-8}{10^{2005}}+\frac{-7}{10^{2006}}\)
Vì \(\frac{-7}{10^{2005}}=\frac{-7}{10^{2005}},\frac{-7}{10^{2006}}=\frac{-7}{10^{2006}},\frac{-8}{10^{2006}}>\frac{-8}{10^{2005}}\)
=) \(\frac{-7}{10^{2005}}+\frac{-7}{10^{2006}}+\frac{-8}{10^{2006}}>\frac{-7}{10^{2005}}+\frac{-8}{10^{2005}}+\frac{-7}{10^{2006}}\)
=) A > B
a) Ta có: 3124= (34)31= 8131
493= (43)31= 64 31
Do 8131 > 64 31 => 3124 < 493
Mà 3124< 3125 => 3125 > 493
3976<42005
\(3< 4;976< 2005\Leftrightarrow3^{976}< 4^{2005}\)