Given that X and y satisfy x-by = 1, y-ax = 1, BX + ay = 1 at the same time, a ^ 2 + B ^ 2 + AB + A + B = 1 is proved

Given that X and y satisfy x-by = 1, y-ax = 1, BX + ay = 1 at the same time, a ^ 2 + B ^ 2 + AB + A + B = 1 is proved

The first and second formulas are expressed by a and B, that is, X and y are solved as unknowns, and we get: x = (1 + b) / 1-ab; y = (1 + a) / 1-ab. then we change X and Y in the third formula into these two formulas, and we can simplify the proof, so we can easily calculate them. Come on!
Given x-by = y-ax = BX + ay = 1, find a ^ 2 + B ^ 2 + AB + A + B
Given a, B, x, y ∈ R +, and ab = 4, x + y = 1, prove (AX + by) (ay + BX) > = 4
(ax+by)(ay+bx)=[(√ax)^2+(√by)^2][(√bx)^2+(√ay)^2]
>=[(√ ax * √ BX + √ by * √ ay] ^ 2 (Cauchy inequality)
=[(√ab)x+(√ab)y]^2 =4(x+y)^2=4
That is, (AX + by) (ay + BX) > = 4, when a = B, take "=" to prove
Hope to help you!
Let the function y = f (x) be defined on the set of real numbers, then what is the image symmetry axis of the functions y = f (x-1) and y = f (1-x)?
The solution shifts the image of F (x) one unit to the right and obtains the image of function y = f (x-1)
The function f (- x) is obtained from the image of F (x) which is symmetric about the y-axis. The function y = f (- x) is translated one unit to the left to get the function y = f (- (x-1)) = f (1-x)
In conclusion, the image symmetry axis of the functions y = f (x-1) and y = f (1-x) is x = 0
It's the y-axis
Given that the equation / X / = ax + 1 about X has a negative root, but no positive root, what is the value range of the real number a? (specific problem solving process)
When x is greater than 0, the above formula can be reduced to: x = ax + 1, that is, X (1-A) = 1, then x = 1 / (1-A). The title knows that x is less than 0, so 1 / (1-A) is less than 0, then a > 1
When x is less than 0, the above formula can be changed to: - x = ax + 1, that is, X (1 + a) = - 1 to get x = - 1 / (1 + a). The title knows that x is less than 0, so - 1 / (1 + a) is less than 0 to get a > - 1, and the same big is taken as the big, so a > 1
X = ax + 1, i.e. a = 1-1 / x,
If x > 0 has no solution and a ≠ 1-1 / x, then a ≥ 1
There is no solution for x = 0
If x < 0 has a solution and a = 1-1 / x, then a > 1
In conclusion, a > 1
What is the equation of the symmetry axis of the image with the function y = cos (2x + 1 / 2)?
As the title And what is the center of symmetry
The equation of axis of symmetry is cos (2x + 1 / 2) = ± 1
2x+1/2=k∏
x=k∏/2-1/4
The center of symmetry is cos (2x + 1 / 2) = 0
2x+1/2= ∏/2+k∏
x=-1/4+∏/4+k∏/2
1) From 2 (x + 1 / 4) to - 1 / 4 + K Π / 2
2) - 1 / 4 - Π / 4 + K Π / 2 or Π / 4-1 / 4 + K Π / 2
1. Axis of symmetry. 2X + 1 / 2 = k π, that is, x = (K π) / 2-1 / 4, where k is an integer;
2. Center of symmetry. If x = (K π) / 2 + π / 4-1 / 4, then the center of symmetry is [(K π) / 2 + π / 4-1 / 4, 0], where k is an integer.
Let 2x + 0.5 be equal to 3.1415926 of N times
It is known that the equation LXL = ax + 1 has a negative root but no positive root
There is a negative root, | x | = - X
-x=ax+1,x=-1/(a+1)0,a>-1.
x> 0 has no solution
x=ax+1
(1-a)x=1
So a = 1
x-1.
No problem. It's fast up there
A symmetric axis equation of the function y = cos (2x - π / 4) is (?)
2X - π / 4 = k π why = k π
X = π / 8 + K π / 2, K is an integer
k=0,x=π/8,C
Is that my answer?
It must be right
There are many symmetrical axes of cosx. The straight line perpendicular to the X axis passing through the highest or lowest point is his symmetrical axis, that is, the symmetrical axis is x = k π
The period of COS (x) is 2 π,
So the period of COS (2x - π / 4) is π.
What is the equation of axis of symmetry for y = cosx?
X = k π, right
So take (2x - π / 4) as a whole
Its axis of symmetry is also K π
The axis of symmetry can pass either the peak or the valley
Because its period is π
If the proposition "∃ x ∈ R, such that x2 + (A-1) x + 1 < 0" is a false proposition, then the value range of real number a is ()
A. 1≤a≤3B. -1≤a≤1C. -3≤a≤3D. -1≤a≤3
∃ proposition "∃ x ∈ R, so that x2 + (A-1) x + 1 ＜ 0" is a false proposition, ∀ x ∈ R, so that x2 + (A-1) x + 1 ≥ 0, ∀ Δ= (A-1) 2-4 ≤ 0, ∀ 1 ≤ a ≤ 3
What is the equation of the axis of symmetry of the graph with the function y = cos (2x + 90 degrees)?
I know it's K π
But why not 2K π
The largest interval of cosx is not 2K π
2X + 90 degrees = k π
So x = k π / 2 - π / 4
x=(2k-1)π/4
Where k is an integer
(of course, there is 2K π in K π. Now that you know it, you can calculate it with K π. K π is its axis of symmetry.)
The function y = cosx, the symmetry axis of image is a straight line x = k π (k is an integer), let 2x + π / 2 = k π. = = = > x = (2k-1) π / 4. = = =, the equation of symmetry axis is: x = (2k-1) π / 4
First, the induced formula is reduced to - SiNx
You have to think of the whole 2x in brackets as an X
Then, because the function y = SiNx, the symmetry axis of the image is a straight line x = k π + π / 2 (k is an integer)
Let 2x = π / 2 + K π. Deduce x = k π / 2 + π / 4
Why is it hard to say, just remember how to remember
First find one that meets the condition, and then look at the next one that meets the condition. What's the difference between them, and then multiply by K
For example, the axis of symmetry of sin, the first one is x = π / 2, and the next one is expanded
First, the induced formula is reduced to - SiNx
You have to think of the whole 2x in brackets as an X
Then, because the function y = SiNx, the symmetry axis of the image is a straight line x = k π + π / 2 (k is an integer)
Let 2x = π / 2 + K π. Deduce x = k π / 2 + π / 4
Why is it hard to say, just remember how to remember
First find one that meets the condition, and then look at the next one that meets the condition. What's the difference between them, and then multiply by K
For example, the axis of symmetry of sin, the first is x = π / 2, the next is x = 3 π / 2, the difference between the two is π, so it is k π, and the center of symmetry is the same
For example, the maximum value, also take sine as an example, the first maximum value is π / 2, the next maximum value is 5 π / 2, the difference between the two is 2 π, so it is 2K π
There are also increasing and decreasing intervals. The first increasing interval of sine is [- π / 2, π / 2], the next is [3 π / 2,5 π / 2], and the difference between them is 2 π, so it is 2K π