solve x#y = (x^x + x^x)^2 - 2(x^x + x^x)(y^y + y^y) + (y^y + y^y)?

Question

if x#y = (x^x + x^x)^2 - 2(x^x + x^x)(y^y + y^y) + (y^y + y^y)
then what is the the ans. of 5#5

Answer 1

x#y = (x^x + x^x)^2 - 2(x^x + x^x)(y^y + y^y) + (y^y + y^y)
5#5 = (5^5+5^5)^2 - 2(5^5+5^5)(5^5+5^5) + (5^5+5^5)
= (3125 + 3125)^2 - 2(3125+3125)(3125+3125) + (3125+3125)
= (6250)^2 - 2(6250)(6250) + (6250)
= 39062500 - 2(39062500) + 6250
= 39062500 - 78125000 +6250
= - 39056249
Ans is 5#5 = - 39056249

Answer 2

If '(y^y + y^y)' is squared in the end of expression, then the 5#5 will be 0.

Else,
As, x#y = (x^x + x^x)^2 - 2(x^x + x^x)(y^y + y^y) + (y^y + y^y)

5#5 = (5^5 + 5^5)^2 - 2(5^5 + 5^5)(5^5 + 5^5) + (5^5 + 5^5)
= (2*5^5)^2 - 2(2*5^5)(2*5^5) + 2*5^5
= 4*5^5*5^5 - 8*5^5*5^5 + 2*5^5
= 2*5^5(2*5^5 - 4*5^5 + 1)
= 2*5^5(6250 - 12500 + 1)
= 2*5^5(-6249)
= (6250)*(-6249)
= - 39056250

Answer 3

If the end of the x#y expression is supposed to be (y^y + y^y)^2, then x#y simplifies to (x^x + x^x - y^y + y^y)^2. So 5#5 would be zero.

Otherwise, you have:
(5^5 + 5^5)^2 - 2(5^5 + 5^5)(5^5 + 5^5) + (5^5 + 5^5) =
(5^5 + 5^5)^2 - 2(5^5 + 5^5)^2 + (5^5 + 5^5) =
-(5^5 + 5^5)^2 + (5^5 + 5^5) =
(5^5 + 5^5)(1 - (5^5 + 5^5)) =
2(5^5)(1 - 2(5^5)) =
6250(1 - 6250) = -39056250

Answer 4

You must have at least ONE X chromosome. It has many many more genes than the Y which are vital and not located elsewhere. So YY=nonviable Many people here are saying YY is an impossibility because no combination of normal sperm or normal egg can give you a YY. This is true, but the question asks "what if" you end up with a YY... And the answer is that it would not be viable because you are missing too many genes that are necessary for development that are located on the X which may have NOTHING to do with sex determination.

Answer 5

If the above equation supposed to have (y^y + y^y)^2 in the end, then
x#y = [(x^x + x^x) - (y^y + y^y)]^2. This is similar to the following algebraic equation:
(a-b)^2 = a^2 - 2ab - b^2, when 'a' is not equal to 'b'.
Since the question here (5#5) equates a to b, the above formula cannot be followed in this case.

Answer 6

if (x^x + x^x) = A and (y^y + y^y) = B
then the equation become
x#y = A^2 -2AB + B^2
= (A-B)^2
= (2x^x-2y^y)^2
= 4(x^x-y^y)^2

so 5#5 = 4(5^5-5^5)^2
= 0


hope can help u
thanks

Answer 7

x#y = (x^x + x^x)^2 - 2(x^x + x^x)(y^y + y^y) + (y^y + y^y)

I bet you miss typed ^2 at the last line, so
x#y = (x^x + x^x)^2 - 2(x^x + x^x)(y^y + y^y) + (y^y + y^y)^2
x#y = ((x^x+x^x)-(y^y+y^y))^2
x#y = (2x^x-2y^y)^2
x#y = 2 (x^x-y^y)^2

for x = 5 and y = 5,
5#5 gives:
x#y = 2 (5^5-5^5)^2
= 2(0)^2
= 0

****** But if you didn't misstype it, then it is so hard to be factorized or solved 5#5, you need MATLAB then

Answer 8

if (x^x + x^x) and (y^y + y^y)

x#y = (x^x + x^x)^2 -2(x^x + x^x)(y^y + y^y) + (y^y + y^y)^2
= ((x^x + x^x)-(y^y + y^y))^2
= (2x^x-2y^y)^2
= 4(x^x-y^y)^2

so 5#5 = 4(5^5-5^5)^2
= 0

Answer 9

If x and y are the same, then

x#x = (x^x + x^x)^2 - 2(x^x + x^x)(x^x + x^x) + (x^x + x^x)
x#x = (2x^x)^2 - 2(2x^x)(2x^x) + (2x^x)
x#x = 4x^(2x) - 8x^(2x) + 2x^x
x#x = -4x^(2x) + 2x^x
so,
5#5 = -4(5^10) + 2(5^5)
5#5 = -4(9,765,625) + 2(3,125)
5#5 = -4(9,765,625) + 2(3,125)
5#5 = -39,056,250

Answer 10

(x^x + x^x)^2 - 2(x^x + x^x)(y^y + y^y) + (y^y + y^y) =
= [(x^x + x^x) - (y^y + y^y)]^2

according to (a - b)^2 = a^2 - 2ab + b^2 where

a = x^x + x^x
b = y^y + y^y

Now, take x=y = 5

[(x^x + x^x) - (y^y + y^y)]^2 = [(x^x + x^x) - (x^x + x^x)]^2 =
= 0^2 = 0