Given that real numbers a, B, x, y satisfy a + B = 9, x + y = 25, then what is the maximum value of AX + by? There must be a process!

Given that real numbers a, B, x, y satisfy a + B = 9, x + y = 25, then what is the maximum value of AX + by? There must be a process!

Let a = 3cosa, B = 3sina
x=5cosB y=5sinB
ax+by=15cosAcosB+15sinAsinB
=15cos(A-B)
So max = 15
If you've studied Cauchy inequality
Then (a ^ 2 + B ^ 2) (x ^ 2 + y ^ 2) > = (AX + by) ^ 2
So (AX + by) ^ 2 = (AX + by) ^ 2
So (AX + by) ^ 2 = (AX + by) ^ 2
So (AX + by) ^ 2
High one mathematics: known real number ABXY satisfy a square + b square = m, x square + y square = n, find the maximum value of AX + by!
ax+by≤（a^2+x^2）/2+(b^2+y^2)/2=(m+n)/2
If and only if a = x and B = y, take the equal sign, the maximum value is (M + n) / 2
Given that the real numbers a, B, x, y satisfy a + B = 1, x + y = 3, then the maximum value of AX + BX is
A + B = 1 Center (0,0) radius 1
X + y = 3 radius of center (0,0) root 3
When a = b = 1, x = y = root 3, the maximum value of AX + by is 2 root 3
Or a = b = - 1, x = y = - radical 3, the maximum value of AX + by is 2 radical 3
Introducing parameters...
Let a = Sina, B = cosa
X = root 3sinb y = root 3cosb
In this way, ax + BX = radical 3 (sinasinb + cosacosb) = radical 3 × (sin (a + b))
So the maximum is the root 3
Given that the real number ABC satisfies √ (a ^ 2 + 2) + | B + 1 | + (c + 3) ^ 2 = 0, find the root of the equation AX ^ 2 + BX + C = 0
The root, absolute value and square are greater than or equal to 0
If one is greater than 0, then at least one is less than 0
So all three equations are equal to zero
So a & sup2; + 2 = 0
A & sup2; = - 2, not true
So there is no solution
Title (topic)
∴a^2+2=0
b+1=0
c+3=0
A = √ - 2 (no, it's wrong, how can it be negative???)
b=-1
c=-3
If the function f (x) = 2cos (ψ x + φ) + m has a straight line x = π / 8 and f (π / 8) = - 1, then the value of real number m is equal to?
For the function y = cosx,
Its axis of symmetry is the x value corresponding to the extreme value of the function!
That is to say, when x is on the axis of symmetry, the function y = 1 or - 1
Therefore:
2+m=-1
Or:
-2+m=-1
The solution is as follows
M = - 3 or 1
For the function y = cosx, the axis of symmetry is the x value corresponding to the extreme value of the function! In other words, when x is on the axis of symmetry, the function y = 1 or - 1, so: 2 + M = - 1 or - 2 + M = - 1 is the best solution
Function y = cosx, its axis of symmetry is the corresponding x value when the function takes the extremum! When x is on the axis of symmetry, the function y = 1 or - 1, so the solution of 2 + M = - 1 or - 2 + M = - 1 is m = - 3 or 1
Given that real numbers a and B satisfy the equations a2-7a + 2 = 0 and b2-7b + 2 = 0, then a / B + B / a =?
If the real numbers a and B satisfy the equations a2-7a + 2 = 0 and b2-7b + 2 = 0, then a / B + B / a =?
Since the real numbers a and B satisfy the equations a2-7a + 2 = 0 and b2-7b + 2 = 0, a and B are two parts of the equation x2-7x + 2 = 0
So a + B = 7, ab = 2,
a/b+b/a=（a2+b2)/ab={(a+b)2-2ab}/ab
={49-4}/2=45/2
Note: A2, B2, (a + b) 2 are square of a, B and (a + b) respectively
The real numbers a and B satisfy the conditions a2-7a + 2 = 0, b2-7b + 2 = 0,
We can regard a and B as two roots of the equation x2-7x + 2 = 0,
∴a+b=7，ab=2，
∴ ba+ab= a2+b2ab= (a+b)2-2abab
= 49-42= 452．
Given that f (x) = 2cos (ω x + θ) + B holds f (x + π / 4) = f (- x) for any real number x, and f (π / 8) = - 1, find the value of B?
It is known that the function f (x) = 2cos (ω x + θ) + B holds f (x + π / 4) = f (- x) for any real number x, and f (π / 8) = - 1,
Find the value of B?
Since f (x + π / 4) = f (- x), the,
Let f (x) = f (x - π / 4 + π / 4) = f (π / 4-x)
Thus f (x) is symmetric with respect to the line x = π / 8
The maximum or minimum value of F (x) is taken at x = π / 8,
And f (π / 8) = - 1,
So B + 2 = - 1 or B-2 = - 1
Get b = - 3 or B = 1
Given that a and B are not all zero real numbers, then the root of the equation x2 + (a + b) x + A2 + B2 = 0 is ()
A. There are two negative roots B. There are two positive roots C. There are two real roots with different signs d. There are no real roots
A = (a + b) 2-4 (A2 + B2) = - 3a2 + 2ba-3b2, because a and B are real numbers that are not all zero, if B = 0, then a ≠ 0, and △ = - 3a2 ＜ 0, the original equation has no real root; if B ≠ 0, if △ is regarded as a quadratic function of a, the opening is downward, △ ′ = 4b2-4 × (- 3) × (- 3B2) = - 32b2 ＜ 0, then △ is always less than 0, the original equation has no real root; therefore, the original equation has no real root
The function f (x) is defined on R, and for any real number x, y, f (x + y) = f (x) + F (y) + 2XY + 3 holds, and f (- 1) = 0
(1) Finding the solution of F (1) and f (2)
(2) If the function y = f (x + 1) is even, find the analytic expression of F (x)
Let x = y = 0, f (0) = f (0) + F (0) + 3, f (0) = - 3
Let x = 1, y = - 1, f (0) = f (1) + F (- 1) - 2 + 3, f (1) = - 3-1 = - 4
Let x = y = 1, f (2) = f (1) + F (1) + 2 + 3, f (2) = - 4-4 + 2 + 3 = - 3
Y = f (x + 1) is an even function, that is, y = f (x + 1) is symmetric about the Y axis, which is obtained by y = f (x) moving 1 to the left
So y = f (x) is symmetric with respect to x = 1, so f (1 + x) = f (1-x)
Substituting into the original formula, we get f (1) + F (x) + 2x + 3 = f (1) + F (- x) - 2x + 3
f(x)=f(-x)-4x
Let y = - x, f (0) = f (x) + F (- x) - 2x & sup2; + 3, then f (- x) = - f (x) + 2x & sup2; - 3
Combined with the above formula, f (x) = - f (x) + 2x & sup2; - 6-4x is obtained
So f (x) = x & sup2; - 2x-3
① f(x+y)=f(x)+f(y)+2xy+3
So f (1 + 1) = f (1) + F (1) + 2 + 3 and f (- 1 + 2) = f (- 1) + F (Y-2) + 2 * (- 1) * 2 + 3, f (2) = - 3, f (1) = - 4
② f(2) = f(x+1+(-x+1)) = f(x+1) + f(-x+1) +2(x+1)(-x+1) +3 = 2f(x+1) -2x^2 + 5
Given that the equation | x | = ax + 1 has a negative root but no positive root, the value range of a is ()
A. A ≥ 1b. A ＜ 1C. - 1 ＜ a ＜ 1D. A ＞ - 1 and a ≠ 0
∵ the equation | x | = ax + 1 has a negative root but no positive root, ∵ x ＜ 0, the equation is changed to: - x = ax + 1, X (a + 1) = - 1, x = − 1A + 1 ＜ 0, ∵ a + 1 ＞ 0, ∵ a ＞ - 1, and a ≠ 0. If x ＞ 0, | x | = x, x = ax + 1, x = 11 − a ＞ 0, then 1-A ＞ 0, the solution is a ＜ 1. ∵ a ＜ 1 is not established without positive root, ∵ a ＞ 1. Therefore, select a