Given that ABXY belongs to R A ^ 2 + B ^ 2 = 1 x ^ 2 + y ^ 2 = 4, find the maximum value of AX + by

Given that ABXY belongs to R A ^ 2 + B ^ 2 = 1 x ^ 2 + y ^ 2 = 4, find the maximum value of AX + by

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（a^2+b^2）（x^2+y^2）=4≥ （ax+by）^2
-2≤ax+by≤2
The maximum value should be 2
Is the answer wrong?
If f (x) = x square + ax + 3 - 1 is less than or equal to 1, the minimum value is - 3
Is the minimum value of - 3 in the interval - 1 less than or equal to 1
f(x)=(x+a/2)²-a²/4+3
-1
Given that the function f (x) = (M2 + m + 1) XM2 − 2m − 1 is a power function and even function, the value of real number m is______ .
∵ function f (x) = (M2 + m + 1) x & nbsp; M2 − 2m − 1 is a power function ∵ we can get M2 + m + 1 = 1, and the solution is m = - 1 or 0. When m = - 1, the function is f (x) = X2, which is even function. When m = 0, the function is f (x) = X-1, which is odd function, not even function in its domain of definition, which is not sufficient. So the answer is: - 1
The solution of inequality 56x square + ax-a square 0 about X
(1) . solve equation 56x ^ 2 + ax-a ^ 2 = 0
It can be divided into: (7x + a) (8x-a) = 0, two are: X (1) = - A / 7, X (2) = A / 8
The original inequality is decomposed into (x + A / 7) (x-a / 8) 0, - A / 7
If the power function f (x) = (m ^ 2-m-1) x ^ m ^ 2-2m-1 is an increasing function in the interval (0, + ∞), find the value of real number M
Power function f (x) = (m ^ 2-m-1) x ^ (m ^ 2-2m-1)
m^2-m-1=1
m=-1
M=2
When m = 2
m^2-2m-1=-1
Power function f (x) = 1 / X
It is a decreasing function in the interval (0, + ∞)
When m = - 1
m^2-2m-1=2
Power function f (x) = x ^ 2
It is an increasing function in the interval (0, + ∞)
The value of real number m is - 1
If x is less than 3, then the solution set of the inequality ax is greater than 3x + 1
ax ＞ 3x + 1
So (3 - a) x is less than - 1
Because a < 3
So 3 - a > 0
So x ＜ - 1 / (3 - a)
Ax>3x+1
(3-A)x
The function y = (m ^ 2-m-5) x ^ M-1 is a power function. When x > 0, it is an increasing function. Find the value of real number M
The coefficient of power function is M & # 178; - m-5 = 1
(m-3)(m+2)=0
x> The index M-1 is greater than 0 with the increase of 0
So m = 3
jhhiuih
m^2-m-5>0
m-1>0
The solution is m > 1 / 2 + radical 21 / 2
On the inequality ax ^ 2 + a-3x > 0 of X, where x ∈ [0, √ 3] has a range of solutions to a
The inequality ax ^ 2 + a-3x > 0 of X has solutions on X ∈ [0, √ 3]
That is, there exists x ∈ [0, √ 3] which makes the inequality hold
That is a (x ^ 2 + 1) > 3x
a> 3x / (x ^ 2 + 1) set up
When x = 0, a > 0 is needed
When 00 infinitely approaches 0
∴a>0
If we define the operation a ⊗ B = B, a ≥ Ba, a < B, then the range of function f (x) = x ⊗ (2-x) is______ .
From a ⊗ B = B, a ≥ Ba, a < B, f (x) = x ⊗ (2-x) = 2 − x, X ≥ 1 x, x < 1, ⊗ f (x) is an increasing function on (- ∞, 1), a decreasing function on [1, + ∞), and ⊗ f (x) ≤ 1, then the range of F (x) is: (- ∞, 1), so the answer is: (- ∞, 1]
What is the solution set of a ^ 27-3x
Like the title,
ax＞3-3x
∴a＞3(1-x)/x (x≠0)
∵a^2