If n is a positive integer, compare the size of 2n times √ (2008-m) and 2n + 1 times √ (m-2009) Any method will do Speed, wealth value is not enough, can only give 5. Thank you!

If n is a positive integer, compare the size of 2n times √ (2008-m) and 2n + 1 times √ (m-2009) Any method will do Speed, wealth value is not enough, can only give 5. Thank you!

According to 2n times √ (2008-m), the number of square root 2008-m ≥ 0, m ≤ 2008;
Then m-2009 < 0, then [2n + 1 times √ (m-2009)] < 0,
Compared with 2n times √ (2008-m) and 2n + 1 times √ (m-2009), the former is larger

Let n be a positive integer: (- 1) 2n (- 1) 2n+
If you're ready,

（-1）2n （-1）2n+1=1*（-1）=-1
Because n is an integer
So 2n is even, so (- 1) 2n = 1
2n + 1 is odd, so (- 1) 2n + 1 = - 1