Let x, y ∈ R, a ＞ 1, B ＞ 1, if AX = by = 3 / 2, a + B = 3, find the maximum value of (1 / x) + (1 / y)?

Let x, y ∈ R, a ＞ 1, B ＞ 1, if AX = by = 3 / 2, a + B = 3, find the maximum value of (1 / x) + (1 / y)?

ax＝by＝3/2→a＝3/(2x),b＝3/(2y).
∴a+b＝3→3/(2x)+3/(2y)＝3.
1 / x + 1 / y = 2 is the fixed value?
Given a, B, X ∈ R, and a ^ 2 + B ^ 2 = 1, x ^ 2 + y ^ 2 = 4, then Max ax + by is no trigonometric ratio!
It can be seen from Cauchy's inequality
(a ^ 2 + B ^ 2) (x ^ 2 + y ^ 2) ≥ (AX + by) ^ 2, i.e
4≥（ax+by）^2
-2≤ax+by≤2
Max ax + by is 2
4=(a^2+b^2)(x^2+y^2)>=（ax+by）^2
So the maximum is 2
Let a = x = 2Sin α
b=y=2cosα
The maximum value is 4
Let a, B, x, y ∈ R and satisfy A2 + B2 = m, X2 + y2 = n, the maximum value of AX + by is ⊙___ .
From Cauchy inequality, we can know that (A2 + B2) (x2 + Y2) ≥ (AX + by) 2, that is, 1 ≥ (AX + by) 2, ax + by ≤ Mn, so the answer is: Mn
If the solution of the equation MX-1 = 2x of X is a negative real number, then the value range of M is
mx-1=2x
(m-2)x=1
x=1/(m-2)
Given that the image of the power function y = x ^ (m ^ 2-2m-3) (m ∈ n *) does not intersect the coordinate axis and is symmetric about the Y axis, then M=____ .
Let f (x) = x ^ (m ^ 2-2m-3)
Because the function image is symmetric about the Y axis
So f (x) = f (- x)
That is, x ^ (m ^ 2-2m-3) = (- x) ^ (m ^ 2-2m-3)
So the value of (m ^ 2-2m-3) should be an even number
Because m is a positive integer and there is no intersection between the function image and the coordinate axis
So the value of (m ^ 2-2m-3) should be a negative even number
So m can only be 1
PS "=" this thing is an equal sign
If the equation | 1-x | = MX has a solution, then the value range of real number M______ .
|1-x | = MX, ① when x ≥ 1, X-1 = MX, (1-m) x = 1, m ≠ 1, x = 11 − m, ｜ 11 − m ≥ 1, the solution is 0 ＜ m ＜ 1; ② when x < 1, 1-x = MX, (1 + m) x = 1, m ≠ - 1, x = 11 + m, 11 + m ＜ 1, ｜ 1 + m ＜ 0 or 1 + m ≥ 1, ｜ m ＜ - 1 or m ≥ 0; in conclusion: the solution set is: m ≥ 0
Given that the image of the power function y = x ^ (m ^ 2-2m-3) (m ∈ z) does not intersect the coordinate axis and is symmetric about the Y axis, try to find the analytic expression of F (x)
With respect to Y-axis symmetry, it's an even function, so the exponent is even
And x-axis, Y-axis have no intersection
Then its image is similar to the inverse scale function in a quadrant
That is, the index is less than 0
So m ^ 2-2m-3
The image does not intersect the coordinate axis,
So M & # 178; - 2m-3
If the equation 22x + 2x · a + 1 = 0 about a has real roots, then the value range of real number a is______ .
Let 2x = t > 0, the original equation is T2 + at + 1 = 0. {a = − T2 − 1t = − t − 1t, t > 0 {a ≤ - 2, if and only if t = 1, the equal sign holds. So the value range of real number a is (- ∞, - 2]. So the answer is: (- ∞, - 2]
If the image of the power function y = x ^ m ^ 2-2m-3 (M belongs to 2) has no common point with the two coordinate axes and is symmetric about the Y axis, then M=
Because m is an integer, G (m) = m ^ 2-2m-3 is an integer
Y = x ^ G has no common point with the two axes and is symmetric about the Y axis, so g (m) is a negative even number
While g (m) = (m-1) ^ 2-4 > = - 4, the possible values are - 4, - 2
When g = - 2, M is not an integer and is rounded off
In this case, y = x ^ (- 4)
One
(17) We know that the solution set of the inequality where the square of ax - 3x + 2 is greater than 0 is {X / X is less than 1 or X is greater than B} 1. Find the value of a and B and solve the square of the inequality ax - (a +...)
(17) It is known that the solution set of the inequality where the square of ax - 3x + 2 is greater than 0 is {X / X is less than 1 or X is greater than B} 1. Find the value of a and B. solve the inequality where the square of ax - (a + b) x + B is less than o,
Substituting x = 1 into the equation AX & # 178; - 3x + 2 = 0, we can get
a-3+2=0
A=1
The inequality is X & # 178; - 3x + 2 > 0
The solution is x2
So B = 2
Two
x²-3x+2
The solution set of ax ^ 2-3x + 2 > 0 is {X / XB}
1 + B = 3 / A, 1 * b = 2 / a (Veda) a = 1, B = 2
(2)ax^2-(a+b)x+b