Let P (x, y) be a moving point on the ellipse with the square of 4 / x + the square of 3 / x = 1 The straight line passing through the point Q (2 pieces of 3,0) to make the inclined angle a intersects with the square of 4 / x + the square of 3 / x = 1 of the ellipse and two different points m, N. find the value range of the absolute value QM multiplied by the absolute value QN

Let x = 2sina, y = 3 ^ cosa
Square of 4 / x + square of 3 / x = 4sin (a + 60) = 1
So the original form of "4-4" is "4"

F (SiNx) = cos3x, then f (cosx) =?

F (SiNx) = cos3x = 4 (cosx) ^ 3-3cosx = cosx [4 (cosx) ^ 2-3] = [(1-sinx) ^ (1 / 2)] * [1-4 (SiNx) ^ 2] so f (cosx) = [(1-cosx) ^ (1 / 2)] * [1-4 (cosx) ^ 2] = SiNx * [1-4 (cosx) ^ 2] ^ is a power * is a multiplier sign, time is in a hurry, I don't know, see you!

It is known that the intersection points of the straight lines y = 2x + 1 and y = 3x + B are in the third quadrant

According to the meaning of the question, we get y = 2x + 1y = 3x + B, the solution is x = 1 − by = 3 − 2B, so the intersection coordinates of y = 2x + 1 and y = 3x + B are (1-B, 3-2b), ∫ the intersection is in the third quadrant, ∫ 1 − B < 03 − 2B < 0, the solution is b > 32, that is, the value range of B is b > 32

Find f (x) = 1 / 3x Λ 3 + 1 / 2x Λ 2-6x + 3

What is the topic? There is only one similar topic on the Internet
The first question is the tangent equation in (0, f (0)), and the second question is the maximum and minimum in [- 3,1]
f(x)=2/3x^3-2x^2-6x+1
f'(x)=2x²-4x-6
f'(0)=-6=k
Tangent point (0,1)
therefore
The tangent is Y-1 = - 6x
Namely
y=-6x+1
(2)
f'(x)=2x²-4x-6=0
x²-2x-3=0
（x+1）（x-3）=0
X = - 1 or x = 3
f(-3)=-17
f(-1)=13/3
f(1)=-19/3
therefore
Maximum = f (- 1) = 13 / 3
Minimum = f (- 3) = - 17

A cube paper box can just put a cylinder with a volume of 6280 cubic centimeters. How big is the volume of the box?
Don't use equations. There's a way. What does 6280 / 157 * 200 mean? Where do 157 and 200 come from

The volume of cylinder is π × radius and height, while height is 2 × radius. So, the volume of square is 2 × radius and height is 2 × radius and height is 2 × radius. The volume of square is 8000 cubic meters

It is known that the set h is all the functions f (x) satisfying the following conditions: there exists a real number x0 in the domain of definition such that f (x0 + 1) = f (x0) + F (1)
It is known that the set h is the whole set of functions f (x) satisfying the following conditions: there exists a real number x0 in the domain of definition such that f (x0 + 1) = f (x0) + F (1) holds. If the function g (x) = LG times a / x square plus one, it belongs to the value range of h for finding real numbers

No, the problem is simple, as long as f (x) = 1 / X is brought into f (x0 + 1) = f (x0) + F (1), that is, 1 / (x0 + 1) = 1 / x0 + 1, after solving the equation, there is no solution to x0, so there is no such x0, so according to the meaning of the problem, the negative power of function f (x) = x does not belong to set H

Please help me solve this problem: given that a and B are opposite to each other, C and D are opposite to each other, the absolute value of X is 5, find CD + A + B - | x | (| x | is the absolute value of x)

I guess you must have the wrong number
c. D is reciprocal
In that case
cd+a+b-|x|
=1+0-5
=-4
Two numbers that are opposite to each other add up to 0
Two reciprocal numbers multiply by one

It is known that f (x) = 1 + log, with 2 as the base and X as the true number (1 ≤ x ≤ 4). Find the maximum value of the function g (x) = [f (x)] ^ 2 + F (2x)

G (x) = [f (x)] ^ 2 + F (2x) = 3 + 3log is based on 2, X is true + (log is based on 2, X is true) ^ 2
=(log takes 2 as the base, X as the real number + 3 / 2) ^ 2 + 3 / 4, log takes 2 as the base, X as the real number, which is greater than - 3 / 2 monotonically increasing
0 ≤ log takes 2 as the base, X as the true number ≤ 2, the minimum value is 3, and the maximum value is 13

It is known that a and B are nonzero vectors, M = a + TB (t ∈ R), if | a | = 1, | a | = 2, if and only if t = 1 / 4, | m |, the minimum value, the angle between a and B
Where A.B is a vector

Let x = the angle between a and B
m= a+tb
|m|^2 = |a|^2 + |b|^2 + 2t|a||b|cosx
t =1/4
|m|^2 = 1+4+cosx = 5 + cosx
min |m|^2 at cosx =-1
min |m| = 2
a. B = 180 degrees

The definition field of function f (x) = LG (4 − x) x − 3 is______ ．

From 4 − x ＞ x − 3 ≠ 0, the solution is: X ＜ 4 and X ≠ 3, so the answer is: {x | x ＜ 4 and X ≠ 3}