2004 /2005*(+2005) /2006 < 2004*(+2005) /2005 +2006

M>N              

\(N=\frac{2004+2005}{2005+2006}=\frac{2004}{2005+2006}+\frac{2005}{2005+2006}\)

\(\text{Vì }\frac{2004}{2005}>\frac{2004}{2005+2006};\frac{2005}{2006}>\frac{2005}{2005+2006}\text{nên:}\)

\(\frac{2004}{2005}+\frac{2005}{2006}>\frac{2004}{2005+2006}+\frac{2005}{2005+2006}\)

Vậy M>N

\(\frac{2004}{2005}>0,5\); \(\frac{2005}{2006}>0,5\)nên \(\frac{2004}{2005}+\frac{2005}{2006}>1\)

\(\frac{2004+2005}{2005+2006}\)có hai số hạng ở tử số nhỏ hơn 2 số hạng ở mẫu số => \(\frac{2004+2005}{2005+2006}\frac{2004+2005}{2005+2006}\)

a)A = B

b)A>B

A>B chúc bạn học tốt

\(A=\frac{2005^{2005}+1}{2005^{2006}+1}\)

\(2005A=\frac{2005^{2006}+2005}{2005^{2006}+1}=\frac{2005^{2006}+1+2004}{2005^{2006}+1}=\frac{2005^{2006}+1}{2005^{2006}+1}+\frac{2004}{2005^{2006}+1}\)

\(B=\frac{2005^{2004}+1}{2005^{2005}+1}\)

\(2005B=\frac{2005^{2005}+2005}{2005^{2005}+1}=\frac{2005^{2005}+1+2004}{2005^{2005}+1}=\frac{2005^{2005}+1}{2005^{2005}+1}+\frac{2004}{2005^{2005}+1}\)

Vì \(\frac{2004}{2005^{2006}+1}

A bé hơn B

5^x + 11^y=26

tìm x,y

\(M=\frac{2004}{2005}+\frac{2005}{2006}\)

\(\Rightarrow M>\frac{2004}{2006+2005}+\frac{2005}{2005+2006}\)

\(\Rightarrow M>\frac{2004+2005}{2005+2006}=N\)

\(\Rightarrow M>N\)

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