\(10A=\frac{2005^{2006}+10}{2005^{2006}+1}\)
\(10B=\frac{2005^{2005}+10}{2005^{2005}+1}\)
Rồi bạn so sánh 10A và 10B là ra.
Ai thấy đúng thì ủng hộ nha !!!, sai thì góp ý cho mink nha
Ta có
A <\(\frac{2005^{2005}+2005}{2005^{2006}+2005}=\frac{2005\left(2005^{2004}+1\right)}{2005\left(2005^{2005}+1\right)}\)=\(\frac{2005^{2004}+1}{2005^{2005}+1}\)
\(\RightarrowĐPCM\)
\(A=\frac{2005.\left(2005^{2005}+1\right)}{2005^{2006}+1}=\frac{2005^{2006}+2005}{2005^{2006}+1}=\frac{2005^{2006}+1+2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}\)
\(B=\frac{2005.\left(2005^{2004}+1\right)}{2005^{2005}+1}=\frac{2005^{2005}+2005}{2005^{2005}+1}=\frac{2005^{2005}+1+2004}{2005^{2005}+1}=1+\frac{2004}{2005^{2005}+1}\)
Vì \(2005^{2005}+1< 2005^{2006}+1\)
\(\Rightarrow\frac{2004}{2005^{2005}+1}>\frac{2004}{2005^{2006}+1}\Rightarrow2005B>2005A\Rightarrow B>A\)
mai thi hạ yên tâm mik sẽ giúp bạn
\(\frac{2005^{2005}+1}{2005^{2004}+1}\)< \(\frac{2005^{2005}+2005}{2005^{2006}+2005}\)= \(\frac{2005.\left(2005^{2004}+1\right)}{2005.\left(2005^{2005}+1\right)}\)= \(\frac{2005^{2004}+1}{2005^{2005}+1}\)( vế B )
vậy A < B
Ta có:
\(A=\frac{2005^{2005}+1}{2005^{2006}+1}\)
\(2005\times A=\frac{2005^{2006}+2005}{2005^{2006}+1}\)
\(2005\times A=\frac{2005^{2006}+1+2004}{2005^{2006}+1}\)
\(2005\times A=1+\frac{2004}{2005^{2006}+1}\)
Ta lại có:
\(B=\frac{2005^{2004}+1}{2005^{2005}+1}\)
\(2005\times B=\frac{2005^{2005}+2005}{2005^{2005}+1}\)
\(2005\times B=\frac{2005^{2005}+1+2004}{2005^{2005}+1}\)
\(2005\times B=1+\frac{2004}{2005^{2005}+1}\)
Nhìn vào trên, suy ra A < B