Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(N=\frac{2004+2005}{2005+2006}=\frac{2004}{2005+2006}+\frac{2005}{2005+2006}\)
\(\text{Vì }\frac{2004}{2005}>\frac{2004}{2005+2006};\frac{2005}{2006}>\frac{2005}{2005+2006}\text{nên:}\)
\(\frac{2004}{2005}+\frac{2005}{2006}>\frac{2004}{2005+2006}+\frac{2005}{2005+2006}\)
Vậy M>N
\(A=\frac{2005^{2005}+1}{2005^{2006}+1}\)
\(2005A=\frac{2005^{2006}+2005}{2005^{2006}+1}=\frac{2005^{2006}+1+2004}{2005^{2006}+1}=\frac{2005^{2006}+1}{2005^{2006}+1}+\frac{2004}{2005^{2006}+1}\)
\(B=\frac{2005^{2004}+1}{2005^{2005}+1}\)
\(2005B=\frac{2005^{2005}+2005}{2005^{2005}+1}=\frac{2005^{2005}+1+2004}{2005^{2005}+1}=\frac{2005^{2005}+1}{2005^{2005}+1}+\frac{2004}{2005^{2005}+1}\)
Vì \(\frac{2004}{2005^{2006}+1}
Giả sử : \(\sqrt{2004}+\sqrt{2006}< 2\sqrt{2005}\)
\(\Leftrightarrow2004+2006+2\sqrt{2004.2006}< 4.2005\)
\(\Leftrightarrow\sqrt{2004.2006}< 2005\Leftrightarrow2004.2006< 2005^2\)
\(\Leftrightarrow\left(2005-1\right)\left(2005+1\right)< 2005^2\)
\(\Leftrightarrow2005^2-1< 2005^2\) . BĐT đúng
Vậy \(\sqrt{2004}+\sqrt{2006}< 2\sqrt{2005}\)
Giả sử : \(\sqrt{2004}+\sqrt{2006}< 2\sqrt{2005}\)
\(\Leftrightarrow2004+2006+2\sqrt{2004.2006}< 4.2005\)
\(\Leftrightarrow\sqrt{2004.2006}< 2005\Leftrightarrow2004.2006< 2005^2\)
\(\Leftrightarrow\left(2005-1\right)\left(2005+1\right)< 2005^2\)
\(\Leftrightarrow2005^2-1< 2005^2.\) BĐT đúng
Vậy \(\sqrt{2004}+\sqrt{2006}< 2\sqrt{2005}\)