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Ta có: B=\(\frac{17^{2009}+1}{17^{2010}+1}\)<1 ( Vì 172009+1< 172010+1 )
Nên B=\(\frac{17^{2009}+1}{17^{2010}+1}\)<\(\frac{17^{2009}+1+16}{17^{2010}+1+16}\)
=\(\frac{17^{2009}+17}{17^{2010}+17}\)
=\(\frac{17\left(17^{2008}+1\right)}{17\left(17^{2009}+1\right)}\)
=\(\frac{17^{2008+1}}{17^{2009}+1}\)=A
Vậy A>B
\(\frac{2^{10}\cdot13+2^{10}\cdot65}{2^8\cdot104}\)
\(=\frac{2^{10}\cdot\left(13+65\right)}{2^8\cdot13\cdot8}\)
\(=\frac{2^{10}\cdot78}{2^8\cdot13\cdot8}\)
\(=\frac{2^{10}\cdot13\cdot2\cdot3}{2^8\cdot13\cdot2\cdot4}\)
\(=\frac{2^2\cdot3}{4}\)
\(=3\)
\(=\frac{2^{10}x\left(13+65\right)}{2^8x104}\)
\(=\frac{2^8x2^2x78}{2^8x104}\)
\(=\frac{4x78}{104}\)
\(=\frac{312}{104}=3\)
b) \(\frac{147}{145}=\frac{145+2}{145}=1+\frac{2}{145},\frac{97}{95}=\frac{95+2}{95}=1+\frac{2}{95}\)
Có \(145>95>0\Leftrightarrow\frac{2}{145}< \frac{2}{95}\Leftrightarrow\frac{147}{145}< \frac{97}{95}\).
g) \(-\frac{39}{47}=\frac{-47+8}{47}=-1+\frac{8}{47},\frac{-43}{51}=\frac{-51+8}{51}=-1+\frac{8}{51}\)
Có \(0< 47< 51\Leftrightarrow\frac{8}{47}>\frac{8}{51}\Leftrightarrow\frac{-39}{47}>-\frac{43}{51}\).
thanks bn