At least one of the equations x ^ 2 + 4mx-4m + 3 = 0, x ^ 2 + (m-1) x + m ^ 2 = 0, x ^ 2 + 2mx-2m = 0 has a real root, and the value range of M is

At least one of the equations x ^ 2 + 4mx-4m + 3 = 0, x ^ 2 + (m-1) x + m ^ 2 = 0, x ^ 2 + 2mx-2m = 0 has a real root, and the value range of M is

When △ = B ^ 2-4ac ≥ 0, there is at least one real root
X ^ 2 + 4mx-4m + 3 = 0
b=4m,a=1,c=-4m+3
(4m)^2-4*(-4m+3)≥0
16m^2+16m-12≥0
4m^2+4m-3≥0
(2m+3)(2m-1)≥0
M ≤ - 3 / 2 or m ≥ 1 / 2
x^2+(m-1)x+m^2=0
Where b = (m-1), a = 1, C = m ^ 2
(m-1)^2-4m^2 ≥0
m^2-2m+1-4m^2≥0
-3m^2-2m+1≥0
3m^2+2m-1≤0
(3m-1)(m+1)≤0
1/3≥m≥-1
x^2+2mx-2m=0
B = 2m, a = 1, C = - 2m
(2m)^2-4*(-2m)≥0
4m^2+8m≥0
m^2+2m≥0
m(m+2)≥0
M ≥ 0 or m ≤ - 2

It is known that the univariate quadratic equation x ^ 2 - (2m + 3) x + 2m + 2 = 0 (M is greater than 0)
Let the two real roots of the equation be X1 and X2 respectively (where X1 is less than x2). If y is a function of M and y = x2-2x1, find the functional relation
Under the above conditions, combined with the image of the function, the answer is: when the value range of the independent variable m meets what conditions, y is less than or equal to M
Don't copy,

∵ X & # 178; - (2m + 3) x + 2m + 2 = (x-1) (x-2m-2) = 0, and ∵ x1

What is the maximum distance from the point P on the ellipse Y & # 178 / 25 + X & # 178 / 9 = 1 to the upper focus?

a=5
b=3
c=4
Maximum = 5 + 4 = 9

As shown in the figure, △ ABC is an equilateral triangle, points D and E are on AB and AC respectively, and F is the intersection of be and CD. It is known that ∠ BFC = 120 °. Verification: ad = CE

It is proved that: ∵ - BFC = 120 °, ∵ - ECF = ∵ BFC - ∵ CEB = 120 ° - ∵ ABC is equilateral triangle, ∵ - EBC = 180 ° - 60 ° - ∵ CEB = 120 ° - CEB, ∵ - ECF = ∵ EBC, that is, ? - DCA = ∵ EBC, and ∵ ABC is equilateral triangle, ∵ - CAD = ? BCE = 60 °, AC = CB ≌ △ ACD ≌

How to solve quadratic equation of one variable by complex number

First, the equation is reduced to the form of a + bi by the operation of complex number, and then a = 0, B = 0 are solved

The known function AXF (x) = B + F (x) (AB is not equal to 0), f (1) = 2, and f (2 + x) = - f (2-x) holds for any X in the domain of definition
Finding the analytic expression of function f (x)
x/2*f(x)=f(x)-1
f(x)=2/(2-x)
How did you get this step

If f (x + 2) = - f (2-x), then f (1) = f (- 1 + 2) = - f (2 + 1) = - f (3) f (1) = 2, and f (3) = - 2 is substituted into AXF (x) = B + F (x), then a * 2 = B + 23a * (- 2) = B-2. The solution is a = 1 / 2, B = - 1 x / 2 * f (x) = f (x) - 1 f (x) * [x / 2-1] = - 1F (x) = - 1 / (X-2 / 2) = > F (x) = 2 / (2-x)

The perimeter of a rectangle is ACM, the width ratio is 3cm, how long is the rectangle

Length + width = a △ 2 = A / 2cm
Long + long-3 = A / 2
2 length = A / 2 + 3
Length = (A / 4 + 3 / 2) cm

In the RL Series AC circuit, if the upper voltage of R is 16V and the upper voltage of L is 12V, the total voltage is?

Root (12 * 12 + 16 * 16) = 20V, the effective value of total voltage is 20V

Set a = {x | x is greater than or equal to - 2 and less than or equal to 7}, B = {x | x is greater than m + 1 and less than 2m-1}. If a and B are equal to a, then the range of M is the same

A ∪ B = a, B is a subset of A,
① If B is contained in a, M + 1 ≥ - 2 and 2m-1 ≤ 7, then: - 3 ≤ m ≤ 4
② M + 1 > 2m-1, that is, M + 1 > 2m-1

In △ ABC, ∠ C = 90 °, CD ⊥ AB, the perpendicular foot is D, BC = 5cm, DC = 4cm, find the length of AC and ab

∵△BCD∽△ABC
∴BD/BC=BC/AB=CD/AC
∵BC=5,DC=4,
In right triangle BCD,
BD²=BC²-CD²
∴BD=3
Substituting the lengths of the three sides into the similar proportions, we get AC = 20 / 3, ab = 25 / 3