If we know that | x | = 5, | y | = 2, and XY > 0, then the value of X-Y is equal to () A. 7 or - 7b. 7 or 3C. 3 or - 3D. - 7 or - 3

If we know that | x | = 5, | y | = 2, and XY > 0, then the value of X-Y is equal to () A. 7 or - 7b. 7 or 3C. 3 or - 3D. - 7 or - 3

∵|x | = 5, |y | = 2, ∵ x = ± 5, y = ± 2, ∵ XY > 0, ∵ when x = 5, y = 2, X-Y = 5-2 = 3; when x = - 5, y = - 2, X-Y = - 5 + 2 = - 3

Given x + y = 2, xy = - 5, then YX + XY=______ ．

YX + xy = x2 + y2xy = (x + y) 2-2xyxy when x + y = 2, xy = - 5, the original formula = 22-2 × (- 5) - 5 = - 145

It is known that the real number XY satisfies, - 4 less than or equal to X-Y less than or equal to - 1, - 1 less than or equal to 4-y greater than or equal to 5
Then the value range of 9x-y is obtained

Let k = 9x-y, then y = 9x-k, and substitute it into the known formula
-4 ≤ X - (9x-k) ≤ - 1, - 1 ≤ 4 - (9x-k) ≤ 5,
That is, 8x-4 ≤ K ≤ 8x-1, 9x-5 ≤ K ≤ 9x + 1,
By drawing a diagram, we know that from k = 8x-4, k = 9x + 1, we can get x = - 5, k = - 44;
From k = 8x-1, k = 9x-5, x = 4, k = 31
The result is - 44 ≤ k = 9x-y ≤ 31

Simplification: 1 / X (x + 1) + 1 / (x + 1) + (x + 2) + 1 / (x + 2) (x + 3) +. + 1 / (x + 2006) (x + 2007)

Use the method of split term cancellation, such as 1 / X (x + 1) = 1 / X-1 / (x + 1), and then use the same method to complete. Finally, 2007 / X (x + 2007)