Question

[trần văn đức]

SO SÁNH

a, C= \(\frac{2005^{2005}+1}{2005^{2006}+1}\)và D = \(\frac{2005^{2004}+1}{2005^{2005}+1}\)

Answer 1

[AGT_KTC4]

nhân cả C và D với 2005 rồi tách ra so sánh

Answer 2

[Xyz OLM]

Ta có : \(2005C=\frac{2005\left(2005^{2005}+1\right)}{2005^{2006}+1}=\frac{2005^{2006}+1+2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}\)

\(2005D=\frac{2005\left(2005^{2004}+1\right)}{2005^{2005}+1}=\frac{2005^{2005}+1+2004}{2005^{2005}+1}=1+\frac{2004}{2005^{2005}+1}\)

Vì \(\frac{2004}{2005^{2006}+1}< \frac{2004}{2005^{2005}+1}\Rightarrow1+\frac{2004}{2005^{2006}+1}< 1+\frac{2004}{2005^{2005}+1}\)

=> 2005.C < 2005.D

=> C < D