If the real number x, y satisfies x ^ 2 + y ^ 2-2x-14y + 45 = 0, then the maximum value of Y-3 / x + 2 is

If the real number x, y satisfies x ^ 2 + y ^ 2-2x-14y + 45 = 0, then the maximum value of Y-3 / x + 2 is

If real numbers x and y satisfy x ^ 2 + y ^ 2-2x-14y + 45 = 0, then the maximum value of Y-3 / x + 2 is 0
x^2+y^2-2x-14y+45=0
(x-1)^2+(y-7)^2=5
Let Y-3 / x + 2 = K
y=kx+2k+3
When x = 1, K is the largest
Substituting for K value
K = (7 + radical 5) / 3
The maximum value of Y-3 / x + 2 is (7 + radical 5) / 3

Let a = {y | y = x2 + 2x + A, X ∈ r}, B = {x | 3-x ≤ 0}, if a ⊆ B, then the value range of real number a is______ ．

B = {x | x ≥ 3} from set B, a = {y | y = (x + 1) 2 + A-1, X ∈ r}, a = {y | y ≥ A-1}, ∫ a ⊆ B, ∫ A-1 ≥ 3 ∫ a ≥ 4 from set a. the value range of real number a is [4, + ∞)

Let there be only one element in the set {x | 0 ≤ x ^ 2 + ax + 5 ≤ 4, X ∈ r}, then the value of real number a is the solution to the problem

From 0 ≤ x ^ 2 + ax + 5 ≤ 4,
It is concluded that 0 ≤ (x + A / 2) ^ 2-A ^ 2 / 4 + 5 ≤ 4
The equation (x + A / 2) ^ 2-A ^ 2 / 4 + 5 ≤ 4 and (x + A / 2) ^ 2-A ^ 2 / 4 + 5 ≥ 0 are obtained
From the equation (x + A / 2) ^ 2-A ^ 2 / 4 + 5 ≤ 4, a ^ 2 / 4 ≥ 1 is obtained
Because there is only one subset in the set
So when x = 1, a = - 2
When x = - 1, a = 2
So a = 2 or a = - 2

If a = {x | 0 ≤ X & sup2; + ax + 5 ≤ 4} is a set of single elements, then the value range of real number a is

If the function y = x2 + ax + 5 opens up and the set is a single element set, then as long as x2 + ax + 5 = 4 has a single solution, X2 + ax + 1 = 0 has a single solution, a * A-4 = 0, a = + 2 or - 2

It is stipulated that an operation a ⁃ B = AB + A-B, where a and B are real numbers, then a ⁃ B + (B-A) ⁃ B equals ()
A. a2-bB. b2-bC. b2D. b2-a

A ∧ B + (B-A) ∧ B, = AB + A-B + (B-A) × B + (B-A) - B, = AB + A-B + b2-ab + b-a-b, = b2-b

If the absolute value of inequality x-4 + (x-3) is less than the solution set of a, then the value range of real number a is the sum of nonempty sets

|x-4|+|x-3|≥|(x-4)-(x-3)|=1
So the minimum value of | x-4 | + | x-3 | is 1
|X-4 | + | x-3 | min 1

Decomposition factor: 4A ^ 2B + 10ab ^ 2

:4a^2b+10ab^2
=2ab(2a+5b)

-4a^3b^2+6a^2b-2ab
In addition: (3x + 2) ^ 2-12 (3x-1)

-4a^3b^2+6a^2b-2ab
=-2ab(2a^2-3a+1)=-2ab(2a-1)(a-1)
（3x+2)^2-12(3x-1)
=（3x+2)^2-12(3x+2-3)
=（3x+2)^2-12(3x+2)+36
=(3x+2-6)^2
=(3x-4)^2

Given a ≠ 0, B ≠ 0, 1 / A-1 / b = 3, find 4a-3ab-4b / A + 2ab-b

∵1/a-1/b=3
The original formula = (4 / b-3-4 / a) / (1 / B + 2-1 / a)
=[-4(1/a-1/b)-3]/[-(1/a-1/b)+2]
=(-12-3)/(-3+2)
=15

4A ^ 3B ^ 2-10a ^ 2B ^ 3C find the common factors and decompose them

4a^3b^2-10a^2b^3c
=2a^2b^2(2a-5bc)