x^2y-6xy^2+9y^3

x^2y-6xy^2+9y^3

x^2y-6xy^2+9y^3
=y(x²-6xy+9y²)
=y（x-3y）²
First put forward, then use formula factorization

x^2+6xy+9y^2+3x+9y-4=0
Why can we factorize it into two linear equations

(x+3y)^2+3(x+3y)=0
(x+3y)(x+3y+3)=0
So we have X + 3Y = 0 or x + 3Y + 3 = 0
They are linear equations: y = - X / 3, y = - (x + 3) / 3

Given the line AB (a > b), can you draw a line so that it equals a + B? What about A-B?

Draw a ray, use a compass to measure the length of a, take the end point of the ray as the center, take the length of a as the radius, draw an arc on the ray, and intersect with the ray at a point. Suppose it is a, then use a compass to measure the length of B, take a as the center, take the length of B as the radius, draw an arc on the ray in turn, and have an intersection with the ray

If the intersection coordinates of the linear function y = 2x + 6 and X axis are (- 3,0), then the solution set of the unary linear inequality 2x + 6 ＞ 0 should be () a, X ＞ - 3b, X ≥ - 3
C、x＜-3 D、x≤-3

2X + 6 ＞ 0 → y ＞ 0 → that is, the image of function y = 2x + 6 is above the x-axis
The image of linear function y = 2x + 6 extends upward from left to right. When x ＞ - 3, the ordinate of the point on the image, that is, the value of Y, that is, the value of 2x + B, is greater than 0
Therefore, the solution set of one variable linear inequality 2x + 6 ＞ 0 should be: X ＞ - 3

Let x, x 1 and x 2 be the lengths of the three sides of an obtuse triangle, and find the value range of the real number X

Firstly, we don't consider the absolute value of obtuse triangle x1-x2 √ X1 ^ 2 + x2 ^ 2, if x corresponds to the minimum angle X

Given that CN = 3 ^ n - λ (- 2) ^ n (λ is a non-zero integer), try to determine the value of λ so that CN + 1 > CN holds for any n that belongs to N +

Cn+1 - Cn
=2x3^n-3λ(-2)^n>0
When n is an even number, 2x3 ^ n + 3 λ (- 2) ^ n > 0 becomes - 3 λ < 2x3 ^ n / (- 2) ^ n, take n = 2, - 3 λ < 9 / 2, that is, λ > - 3 / 2
When n is odd, 2x3 ^ n + 3 λ (- 2) ^ n > 0 becomes - 3 λ > 2x3 ^ n / (- 2) ^ n, take n = 1, - 3 λ > - 3, that is, λ < 1
In conclusion - 3 / 2 ＜λ＜ 1

Find the partial derivative of Z = (x ^ 2 + y ^ 2) e ^ [(x ^ 2 + y ^ 2) / XY]

Zx=exp(x/y)*exp(y/x)*(x^4 + 2*x^3*y - y^4)/(x^2*y)
Zy=exp(x/y)*exp(y/x)*(- x^4 + 2*x*y^3 + y^4)/(x*y^2)
Where exp (T) = e ^ t
If satisfied, please accept! ^^

There is a pile of go pieces, which can be arranged into six rows and six columns. How many pieces are there in total? How many pieces are there in the outermost layer of go pieces?

6 * 6 = 36, outermost: 6 * 4-4 = 20

1. No matter what the value of K is, the point that the image of quadratic function y = x ^ 2 + (2-k) x + K must pass through is -——
2. If the image of the function y = (A-1) x ^ 2-2x + 1 does not intersect the X axis, then the value range of a is -——
Why "let 1-x = 0"

1. Y = x ^ 2 + (2-k) x + k = x ^ 2 + 2x KX + K (1-x) k = Y-X ^ 2-2x when 1-x = 0 and Y-X ^ 2-2x = 0, it must be true, so x = 1, y = x ^ 2 + 2x = 3, so the points to pass through are (1,3) 2. If a = 1, then y = - 2x + 1, intersect with X axis, which is not suitable for the problem. If a is not equal to 1, is a quadratic function, and does not intersect with X axis, then y is not equal to 0, that is equation (...)

For example, if the inverse scale function y = k of X passes through a (2,1), if y ≤ 1, then the value range of X
The answer is x ≥ 2, X

y=2/x
y