Let f (x) = x3-3x2-9x + m.1 find the monotone interval of F (x). 2 for X equation f (x), there are three real roots, find the value range of real number m!

1）f'(x)=3x^2-6x-9=3(x^2-2x-3)=3(x-3)(x+1)
We get the extreme point x = 3, - 1
Monotone increasing interval: x3
Monotone decreasing interval: - 1

It is known that the equation 3x2-10x + k = 0 about X has a real root, and the value of K is obtained
1. There is a positive root and a negative root. 2. Both roots are less than 2. 3. One is greater than 2 and one is less than 2

It is known that the equation 3x2-10x + k = 0 about X has a real root, and the value of K is obtained
1. There is a positive root and a negative root
Having a positive root and a negative root can be restricted by the following conditions:
Firstly, the discriminant = 100-12k > 0 can guarantee two unequal real roots;
If there are two different signs, then there is the product of two: K / 3

The interval of equation x & # 179; + 2x & # 178; - 1 = 0 with at least one real root is（

xy^2-x^2y x^3y^2-4x

xy^2-x^2y
=xy(y-x)
x^3y^2-4x
=x(x^2y^2-4)
=x(xy-2)(xy+2)

Given 2x-y = 1 / 3, xy = 2, how many steps does it take to find 4x ^ 3y-3x ^ 2Y ^ 2 + XY ^ 3

2X-Y=1/3
XY=2
4X^3y-3x^2y^2+xy^3
= 4X^3y-4x^2y^2+xy^3 + x^2y^2
= xy (4X^2-4xy+y^2) + x^2y^2
= xy(2x-y)^2 + x^2y^2
= 2 * (1/3)^2 + 2^2
= 2/9 +4
= 38/9

Given x + y = - 4, xy = - 12, find the value of Y + 1x + 1 + X + 1y + 1

The original formula = (y + 1) 2 + (x + 1) 2 (x + 1) (y + 1) = x2 + Y2 + 2 (x + y) + 2XY + X + y + 1 = (x + y) 2 − 2XY + 2 (x + y) + 2XY + X + y + 1 ∵ x + y = - 4, xy = - 12, the original formula = 16 + 24 − 8 + 2 − 12 − 4 + 1 = − 3415

Given x ^ 2 + y ^ 2 = 12, x + y = 4, then XY=
There is another question: x ^ 2-x +? = (x -?) ^ 2

(x+y)²=4²
x²+2xy+y²=4²
2xy+12=16
2xy=4
xy=2
x²-x+1/4=(x-1/2)²
If you don't understand this question, you can ask,

There is a calculation problem: "calculate the value of (2x-3xy-2xy) - (x-2x + y) + (- x + 3xy-y), where" x = 1 / 2, y = - 1 ". Student a mistook x = 1 / 2 for x = - 1 / 2, but the calculation result is still correct. What do you say?

Polynomials have nothing to do with the value of X. you can know it by combining the object like terms yourself

[6xy ^ 2 multiplied by (- 2XY) ^ 2-24x ^ 5Y ^ 6 divided by (- 2XY) ^ 2] divided by (- 3xy ^ 2) ^ 2, where x = 2, y = 2003

[6xy ^ 2 * (- 2XY) ^ 2-24x ^ 5Y ^ 6 / (- 2XY) ^ 2] / (- 3xy ^ 2) ^ 2 = [6xy ^ 2 * 4x ^ 2Y ^ 2-24x ^ 5Y ^ 6 / (4x ^ 2Y ^ 2)] / (9x ^ 2Y ^ 4) = (24x ^ 3Y ^ 4-6x ^ 3Y ^ 4) / (9x ^ 2Y ^ 4) = (18x ^ 3Y ^ 4) / (9x ^ 2Y ^ 4) = 2x because x = 2, the original formula = 2x = 4

The first (- 3xy ^ 2) ^ 2 multiplied by 2XY divided by 3x ^ 2Y ^ 5, and the second (X-Y) ^ 5 divided by (Y-X) ^ 3 requires a process

(- 3xy ^ 2) ^ 2 multiplied by 2XY divided by (3x ^ 2Y ^ 5)? = (9x & # 178; y ^ 4) * 2XY / (3x ^ 2Y ^ 5) = 18x ^ (2 + 1) y ^ (4 + 1) / (3x ^ 2Y ^ 5) = 6x ^ (3-2) y ^ (5-5) = 6xy ^ 0 = 6x (X-Y) ^ 5 divided by (Y-X) ^ 3? = - (Y-X) ^ 5 / (Y-X) ^ 3 = - (Y-X) ^ (5-3) = - (Y-X) &# 178

Let f (x) = x3-3x2-9x + m.1 find the monotone interval of F (x). 2 for X equation f (x), there are three real roots, find the value range of real number m!