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Given m = x + y, n = X-Y, try to use M, n to express (X & # 179; + Y & # 179;) &# 178; + (X & # 179; - Y & # 179;) &# 178;);, 1: Given m = x + y, n = X-Y, try to use M, n to express (X & # 179; + Y & # 179;) &# 178; + (X & # 179; - Y & # 179;) &# 178;);,

x＋y＝m x－y＝n ∴ x＝﹙m＋n﹚/2 y＝﹙m－n﹚/2 x·y＝﹙m²－n²﹚/4 ∴ ﹙x³＋y³﹚²＋﹙x³－y³﹚² ＝﹙x＋y﹚²﹙x²－xy＋y²﹚²＋﹙x－y...

Given that point a (a, 3) and point B (2, b) are symmetric about X axis, then a + B=______ ．

∵ point a (a, 3) and point B (2, b) are symmetric about X axis, ∵ a = 2, B = - 3, then a + B = 2-3 = - 1

Given that point a (2x-1, y + 3) and point B (1-y, X-2) are symmetric about the Y axis, the value of X + y can be obtained

Point a (2x-1, y + 3) and point B (1-y, X-2) are symmetric about the Y axis,
So 2x-1 + 1-y = 0, y + 3 = X-2, x = - 5, y = - 10
x+y=15-10=-15.

If the image of the first-order function y = x + 2 intersects the x-axis, and the y-axis is at point a and point B, what is the area of the triangle OAB

The image of the first-order function y = x + 2 intersects the x-axis respectively, and the y-axis is at points a and B,
We can see that a (- 2,0), B (0,2)
S triangle OAB = 1 / 2 * OA * ob = 2

The area of the triangle formed by the image, x-axis and y-axis is 9, which is an analytic expression of the first-order function
.

Let the analytic formula of a function be y = KX + B
∵ image over point (6,0)
∴0=6k+b ∴b=-6k
∴y=kx-6k
When x = 0, the intersection of y = - 6K  line and Y axis is (0, - 6K)
∴S△=1/2×6×│-6k│=18│k│=9
That is, k = 1 / 2 or K = - 1 / 2
‖ B = - 3 or B = 3
The analytic formula of a function is y = 1 / 2x-3 or y = - 1 / 2x + 3

The area formula of the triangle formed by the image of the first-order function and the x-axis and y-axis

The area formula of triangle is as follows:
S=(1/2)*|x|*|y|.
Where | x | -- the intercept of the image (line) of the first-order function on the x-axis and | y | -- the intercept of the line on the y-axis

The image of a function of degree is known to pass through points (2,1) and (1,3). The coordinates of the intersection of the function of degree with x-axis and y-axis are obtained

Hardy invincible let the analytic expression of the first-order function be: y = KX + B ∵ the image of the first-order function passes through the simultaneous equations of points (2,1) and (1,3) ∵ 2K + B = 1K + B = 3, and the solution is k = - 2, B = 5 ∵ the analytic expression of the first-order function is: y = - 2x + 5, if x = 0, then y = 5, that is, if the coordinate of the first-order function and Y axis is (0,5), then y = 0, then - 2x +

The ordinate of the intersection of the image of a function and the line x + y = 2 is 4, and the intersection with the Y axis is (0,1)

The analytic formula of a function is y = KX + B
Substituting y = 4 into x + y = 2
We get x = - 2
X = - 2, y = 4. X = 0, y = 1 substitute y = KX + B
We get {4 = - 2K + B}
1=b
The solution is k = - 1.5, B = 1
The analytic expression of a function is y = - 1.5x + 1

Given the first-order function y = (3a-2) x + 4-b, the intersection of the image and Y axis is below the X axis

The point of intersection with the Y axis is below the X axis
That is, (3a-2) x + 4-b when x = 0, 4 meets the requirements of the topic

Given that the image of the first-order function passes through the point (1.2), and the coordinate product of the intersection of the image and x-axis and y-axis is 9, the analytic expression of the first-order function is obtained

The product of the abscissa of the intersection of the image and the X axis and the ordinate of the Y axis is 9
Coordinates of the intersection of image and X-axis (- B / K, 0)
Coordinates of the intersection of image and Y axis (0, b)
（-b/k）b=9
The analytic formula of a function y = KX + B substitutes the point (1,2) into 2 = K + B
The solution of the equation is k = - 4 or K = - 1, B = 6 or B = 3
Y = - 4x + 6 or y = - x + 3