Known m R, compare the size of m 6 2009 with m 4 m 2 2008

m^6+2009)-(m^4+m^2+2008)(m^6-m^2)-(m^4-1)

m^2(m^4-1)-(m^4-1)

m^2-1)(m^4-1)

m^2-1)(m^2-1)(m^2+1)(m^2-1)²(m^2+1)

When m -1 = 0, m = 1, (m 2-1) (m +1) = 0 in other cases, (m 2-1) (m +1) > 0 So, when m ≠ 1, m 6 + 2009 > m 4 + m 2 + 2008;When m = 1, m 6 + 2009 m 4 + m 2 + 2008

Difference method: m 6 + 2009 - (m 4 + m 2 + 2008) = m 6 - m 4 - m 2 + 1 = m 4 (m 2-1) - (m 2-1) = (m 2 + 1) (m 2-1) 2 0

When m 2 = 1, that is, m = +1 or -1 the two numbers are equal, and the former is greater than the latter in other cases.

n = 2008 squared - 2008 * 2009 + 2009 squared.

2008 squared - 2 * 2008 * 2009 + 2009 square + 2008 * 2009

2008-2009) + 2008*2009m=2008*2009-1

So n-m=2

So n>m

n-m = 2008 squared - 2 * 2008 * 2009 + 2009 square + 1

2008-2009).

So n>m

We know that LNT is an increment function in its entire defined domain.

t1=m^2 t2=m^m

lnt1-lnt2 = 2lnm-mlnm = (2-m) lnm When m>1 lnm >0

When m>2 - m<0

So (1) when m>2 lnt1-lnt2<0(2) when 10

3) When 0< m<1 lnt1-lnt2<0 when m=1, or m=2 they are equal.

Because m is greater than 0, negative numbers are ignored.

Compare the size of the same base to the power of the same base.

The base is the same, and it must be that the larger the exponent, the larger the value.

So when m<2.

m^2>m^m

When m=2.

m^2=m^m

When m=2.

m^2

I forgot. The power function in college math is the x-power of a.

I remember that the intersection point is at the position where the y-axis is 1, i.e., when the base is less than 1 and greater than 0, it is a subtraction function (the larger the exponent, the smaller the function), and when it is greater than 1, it is an increasing function (the larger the exponent, the larger the function).

So first judge m

When 0m (judgment exponent) and the function is a subtraction function, so m 2m m, when m > 2, because m>2 (judgment exponent) and the function is an increasing function, so m 2 when m = 1 or m = 2 m 2==m m

If m (0,1),n (0,1), then logm4 0, logn4 0, from logm4 logn4 0, we get -logm4 -logn4 0, i.e. log1m

4＞log1n

4＞0，1＜1m＜1

n，∴1＞m＞n＞0；

If m 1, n 1, then logm4 0, logn4 0, logm4 logn4, m n 1;

If n 1 m 0, then logm4 0 logn4 is clearly true, in summary, 0 n m, or m n 1, or 0 m 1 n

m 8/7n

Add 2/8/7 m+2 on both sides> 8/7 n+2

There are two possibilities for this question. That is, it is solved in two cases. 1.If m<0, -8 7m ten2 is less than -8 7n ten.

If m is greater than 0, -8 7m dec2 is greater than -8 7n dec2.

Within the range of rational numbers, of course MN Dala.

Take two special values that match the meaning of the topic, and get m+n < mn

(m+n)(m+n)=m*m+2mn+n*n>=mn*mn

By square comparison, then draw conclusions.

Dude, if your title is m=2008*2009-1;n = 2008 square - 2008 * 2009 + 2009 square meters.

Then it goes like this:

n = 2008 square - 2008 * 2009 + 2009 square meters.

2008*(2008-2009)+2009 sq.m.

1*2008+2009 square meters.

2009 + 1 + 2009 sq.

So n》m(* hee-hee......I hope you can understand it!