Given that point a (2x-6,2y + 5) and point B (y, - x) are symmetric about X axis, then the value of X + y is as soon as possible Come to the number, but also say why

Given that point a (2x-6,2y + 5) and point B (y, - x) are symmetric about X axis, then the value of X + y is as soon as possible Come to the number, but also say why

Two points are symmetric about x-axis. 1) abscissa is equal, that is, 2x-6 = y
2) The ordinates are opposite to each other, that is, 2Y + 5 = X
Formula 1 and formula 2 are added on the left and equal to the right to get x + y = 1

1. Given that point a (x, 4-y) and point B (1-y, 2x) are symmetric about y axis, write the value of x power of Y

For Y-axis symmetry, the abscissa is opposite and the ordinate is equal
So x = - (1-y)
That is, X-Y = - 1
4-y=2x
That is, 2x + y = 4
Solving equations
x=1,y=2
X power of y = 1 power of 2 = 2

Given that the coordinates of point a are (2x + Y-3, x-2y), and the coordinates of point B which is symmetrical about the X axis are (x + 3, y-4), then how much is x = and how much is y =

The point is symmetric about the X axis, that is, the two points are above and below the X axis, the X coordinates are the same, and the Y coordinates are opposite to each other. List this relationship, and then solve the binary linear equation. 2x + Y-3 = x + 3, x = 5, we get x-2y = - (y-4) y = 1, that is 2x + Y-3 = 8, x + 3 = 6x-2y = 3, that is - (y-4) = 3, so the two symmetric coordinates are (8,3) and (8, - 3)

If point a (x + 2y-1,2x-y) and point (X-Y, 2 + x) are symmetrical about the Y axis, then what are the coordinates of two points a and B in turn

Because: About Y-axis symmetry,
So: two points y value is equal, x value is opposite
There are: x + 2y-1 = - (X-Y)
2X-Y = 2+X
Solve the equations, x = 1, y = - 1
Substituting points a and B, we get a (- 2,3), B (2,3)

Given x + y = 4, x 2 + y 2 = 14, find the value of X 3y-2 x 2Y 2 + X Y 3

∵ x + y = 4, ∵ (x + y) 2 = 16, ∵ x2 + Y2 + 2XY = 16, and X2 + y2 = 14, ∵ xy = 1, ∵ x3y-2x2y2 + XY3 = XY (x2-2xy + Y2) = 14-2 = 12

1. X ^ 2 + XY + y = 14, y ^ 2 + XY + x = 28, find the value of X + y 2. X ≠ y, x ^ 2 + XY + y ^ 2 3. Factorization of x ^ 4 + 2x ^ 3 + 2x ^ 3 + 2x + 1

【1】 By adding the two formulas, we get: X & # 178; + 2XY + Y & # 178; + X + y = 42 (x + y) & # 178; + (x + y) - 42 = 0 (x + y + 7) (x + y-6) = 0, then: x + y = - 7 or x + y = 6

Given m = (x + y) / (√ X - √ y) - (2 √ XY) / (√ X - √ y), n = (3 √ X-2 √ y) / [(√ x + y) + (√ Y-X)] which is the largest of Mn
y=[√(x-8)]+[√(8-x)]+18

According to the "Question supplement", x = 8, y = 18
√x=2√2,√y=3√2
M＝-√2
N＝0
∴M＜N

We know the relation of M = {(x, y) | x > 1, Y > 1} n = {(x, y) | x + Y > 2 and XY > 1} Mn

M ∈ n, because M can only have x > 1, Y > 1; n has x > 1, Y > 1, and x = 2, y = 1 or x = 1, y = 2; so m ∈ n

If positive real numbers x and y satisfy 2x + y + 6 = XY, what is the minimum value of XY?

∵ positive real numbers x, y, ∵ XY > 0
∴2x+y≥2√(2xy)
∴2x+y+6=xy≥2√(2xy)+6
That is, XY-2 √ 2 * √ (XY) - 6 ≥ 0
If we solve the inequality, we get
√ (XY) ≥ 3 √ 2 (√ (XY) ≤ - √ 2 abandoned)
∴xy≥(3√2)^2=18
The minimum value of XY is 18

How to remember X & # 179; - Y & # 179; (X-Y) (X & # 178; + XY + Y & # 178;)

In the formula of cubic difference, the sign of odd order is the same, but the sign of even order is opposite
(x ^ 3-y ^ 3) and (X-Y) are the same minus sign, even times the opposite is (x ^ 2-xy...)
This locks the symbol:
So is x ^ 3 + y ^ 3
x^3+y^3=(x+y)(x^2-xy+y^2)