If a = x2-2xy + Y2, B = x2 + 2XY + Y2, then A-B = () A. 2x2+2y2B. 2x2-2y2C. 4xyD. -4xy

If a = x2-2xy + Y2, B = x2 + 2XY + Y2, then A-B = () A. 2x2+2y2B. 2x2-2y2C. 4xyD. -4xy

∵ a = x2-2xy + Y2, B = x2 + 2XY + Y2, ∵ A-B = (x2-2xy + Y2) - (x2 + 2XY + Y2) = x2-2xy + y2-x2-2xy-y2 = - 4xy, so D

The second power of X - 2XY + the second power of Y - the second power of Z

The original formula = (X-Y) &# 178; - Z & # 178;
=(x-y+z)(x-y-z)

Why is 4x ^ 2 y ^ 2 + 2xy-x ^ 2-y ^ 2 factorizable? Please help me --

4X^2 Y^2+2XY-X^2-Y^2
=(2XY)^2-(X-y)^2
=(2XY+X-Y)(2XY-X+y)

The sum of the squares of two consecutive even numbers is 1252. What is the sum of these two numbers?

Let the smaller even number be X
x²+(x+2)²=1252
x²+x²+4x+4=1252
x²+2x=624
x²+2x+1=625
(x+1)²=625
x+1=±25
X1 = 24, X2 = - 26 (rounding off)
x+2=26
The sum of these two numbers is 24 + 26 = 50

The product of three numbers is 48 and the sum of squares is 56

Let the tolerance of the arithmetic sequence be D, and the middle term be a, so the other two numbers are A-D. a + D lists the equation system (A-D) * a * (a + D) = 48 (A-D) ^ 2 + A ^ 2 + (a + D) ^ 2 = 56, that is, a ^ 3-A * d ^ 2 = 48, a ^ 2 + 2D ^ 2 = 56, B 2 * a, get 3 * a ^ 3 + 2A * d ^ 2 = 56a, C 1 * 2, get 2 * a ^ 3-2a * D

It is known that an arithmetic sequence has odd terms, in which the sum of technical terms is 56 and the sum of even terms is 49
(1) Find the number of items of the sequence (2) find the middle item of the sequence

Let n items be set, and the tolerance be d.a1 + a3 +... + an = 56a2 + A4 +... + a (n-1) = 49 [A2 + A4 +... + a (n-1)] - [A1 + a3 +... + an] = (a2-a1) + (a4-a3) +... + [a (n-1) - A (n-2)] - an = (n-1) d / 2 - A1 - (n-1) d = - A1 - (n-1) d / 2 = (- 1 / 2) [2A1 + (n-1) D] = (- 1 / 2) (a1 + an) = 49-56 = -

The sum of the squares of three consecutive positive odd numbers equals 251. Find the smaller prime of the number

Let X be the smallest, then x ^ 2 + (x + 2) ^ 2 + (x + 4) ^ 2 = 251 3x ^ 2 + 12x-231 = 0 x ^ 2 + 4x-77 = 0 (x + 11) (X-7) = 0 x = 7

The square difference of two consecutive odd numbers must be zero___ Multiple of
Fill in the maximum number

The difference between the squares of two consecutive odd numbers must be a multiple of
Because the expression of odd number is (2n-1) or (2n + 1)
So let these two consecutive odd numbers be (2n-1), (2n + 1)
(2n+1)^2-(2n-1)^2
=(2n+1+2n-1)(2n+1-2n+1)
=8n

Given that the square difference of two consecutive odd numbers is 2008, the two consecutive odd numbers can be

Let the two continuous odd numbers be 2k-1 and 2K + 1 respectively, and K be a natural number
According to the meaning and known, there are:
(2k+1)^2-(2k-1)^2=2008
(2k+1+2k-1)(2k+1-2k+1)=2008
8k=2008
k=251
Substituting into the set, there are:
2k-1=2×251-1=501
2k+1=2×251+1=503
A: the two consecutive odd numbers are 501 and 503

The difference between the squares of two odd numbers is 32?

Let one odd number be (2n-1), then the other is (2n + 1)
(2n-1)²-(2n+1)²=32
（2n-1+2n+1）（2n-1-2n-1）=32
4n×（-2）=32
-8n=32
n=-4
In fact, n = 4 is OK, because the square difference is 32
Two odd numbers are 7, 9 or - 7, - 9