a)A = B

b)A>B

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

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

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

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

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

Vì \(\frac{2004}{2005^{2006}+1}

A bé hơn B

5^x + 11^y=26

tìm x,y

 Ta có : A=2005^2005+1/2005^2006+1

=>2005A=2005.(2005^2005+1)/2005^2006+1

=>2005A=2005^2006+2005/2005^2006+1

=>2005A=2005^2006+1+2004/2005^2006+1

=>2005A=2005^2006+1/2005^2006+1 + 1/2005^2006+1

=>2005A=1+1/2005^2006+1

 Lại có:B=2005^2004+1/2005^2005+1

=>2005B=2005.(2005^2004+1)/2005^2005+1

=>2005B=2005^2005+2005/2005^2005+1

=>2005B=2005^2005+1+2004/2005^2005+1

=>2005B=2005^2005+1/2005^2005+1 + 1/2005^2005+1

=>2005B=1+1/2005^2005+1

Vì 2006>2005

=>2005^2006>2005^2005

=>2005^2006+1>2005^2005+1

=>1/2005^2006+1<1/2005^2005+1

=>1+1/2005^2006+1<1+1/2005^2005+1

=>2005A<2005B

=>A<B

Vậy A<B

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