Given that the square of (x-3) + the absolute value of Y + 2 = 0, find the value of the square of the algebraic formula 2x + (- x-2xy + 2Y's Square) - 2 (the square of x-xy + the square of 2Y)

Given that the square of (x-3) + the absolute value of Y + 2 = 0, find the value of the square of the algebraic formula 2x + (- x-2xy + 2Y's Square) - 2 (the square of x-xy + the square of 2Y)

The absolute value of the square of (x-3) + y + 2 = 0
x-3=0 x=3
y+2=0 y=-2
The value of the square of the algebraic formula 2x + (- x-2xy + 2Y's Square) - 2 (the square of x-xy + the square of 2Y)
=18+(-3+12+8)-2(9+6+8)
=18+17-46
=35-46
=-11

Given that the square of (x + 2) and the absolute value of X + Y-1 are opposite numbers to each other, we can find the value of the algebraic formula X / Y-Y / x-x's square + Y / XY RT, quick!

The absolute value of (x + 2) ^ 2 + X + Y-1 = 0,
So x = - 2, y = 3
x/y-y/x-(x^2+y^2)/xy
=(x^2-Y^2-x^2-Y^2)/xy
=-2y^2/xy
=-2y/x
=3

Given the absolute value of 3-y + the absolute value of X + y = 0, find the value of the algebraic formula x + Y / XY

|3-y|+|x+y|=0
So we have 3-y = 0 and X + y = 0
Results: y = 3, x = - 3
(x+y)/(xy)=0

X, y = 5, xy = 3 find the algebraic formula x ^ 3y-2x ^ 2Y ^ 2 = XY ^ 3 Solution·····

Xy = 3 x + y = 5 square of both sides x ^ 2 + 2XY + y ^ 2 = 25 x ^ 2 + y ^ 2 = 25-2 * 3 = 19 x ^ 3y-2x ^ 2Y ^ 2 + XY ^ 3 = XY (x ^ 2 + y ^ 2) - 2 (XY) ^ 2 = 3 * 19-2 * 3 ^ 2 = 39

If x, y, m are suitable for the relation, 3x + 5y-3-m + 2x + 3y-m = x + y-2004 + x-y-m under radical 2004-x-y, try to find the arithmetic square root of M-4

∵ x + y-2004 ≥ 0 and 2004-x-y ≥ 0
∴x+y-2004=0
x+y=2004 (1)
ν 3x + 5y-3-m + 2x + 3y-m = 0
∴3x+5y-3-m=0 (2)
2x+3y-m=0 (3)
From (1), (2), (3), M = 2007
∴m-3=2004

Given that | 2x + 3Y + 1| and (3x-2y-18) 2 are opposite numbers to each other, the 2012 power of (- 4x-5y) is_____ .

It is known that (2x-2x + 3Y + 1) and (3x-2y-18-18) are opposite numbers to each other, but | 2x + 3Y + 3Y + 1 | ≥ 0 (3x-2y-18) 2 ≥ 0 | 2x + 3Y + 1y + 1 | + (3x-2y-18) 2 = 0, so | 2x2x + 3Y + 1y + 1 | = (3x-2y-18) is 0, that is 2x + 3Y + 1 = 3x-2x-2y-18 = 0, the solution x = 4Y = - 3-3-4x-5x-55x-55x = 4Y = 3-4x-5x-2x-2x-2y y = - 16 + 15 = - 1 (- 1) ^ 2012 = 1

A polynomial plus - 2x ^ 3 + 4x ^ y + 5Y ^ 3 yields x ^ 3-3x ^ 2Y + 2Y ^ 3 (1) Find this polynomial (2) when x = - 2, find the value of this polynomial When x = - 2 is wrong x=-1/2,y=1

This polynomial
=(x^3-3x^2y+2y^3)-(-2x^3+4x^y+5y^3)
=3x^3-3x^2y-3y^3-4x^y
When x = - 2
Polynomial = - 24-3 * 4 ^ y-3y ^ 3-4 * (- 2) ^ y

Given a polynomial plus - 2x? + 4x? Y + 5Y? To get x? 3x? Y + 2Y? And find the value of the polynomial, where x Given a polynomial plus - 2x? + 4x? Y + 5Y?, we get x? - 3x? Y + 2Y? And find the value of this polynomial, where x = - 2 ′ 1, y = 1

(X³－3X²Y＋2Y³)-(﹣2x³＋4x²y＋5y³)=3x³-7X²Y-3y³
The value of X and y can be brought in
What's the value of X? I don't understand

Add - 2x cube + 4x squared y + 5Y cube to a polynomial to get x cube - 3x squared y + 2Y cube. Calculate the value of this polynomial

(x³-3x²y+2y³)-(-2x³+4x²y+5y³)
=x³-3x²y+2y³+2x³-4x²y-5y³
=3x³-7x²y-3y³

A polynomial plus - 2x3-3x2y + 5y2 gives x3-2x2y + 3y2 (1) Find this polynomial; (2) When x = 1 2, y = 1, find the value of this polynomial

（1）（x3-2x2y+3y2）-（-2x3-3x2y+5y2）
=x3-2x2y+3y2+2x3+3x2y-5y2
=3x3+x2y-2y2，
A: this polynomial is 3x3 + x2y-2y2
(2) When x = - 1
2, y = 1,
3x3+x2y-2y2．
=3×（-1
2）3+（-1
2）2×1-2×12
=-3
8+1
4-2
=-25
8．