Let P1, P2, P3 be three prime numbers, and P2 = P1 + 4, P3 = P1 + 8, prove P1 = 3

Let P1, P2, P3 be three prime numbers, and P2 = P1 + 4, P3 = P1 + 8, prove P1 = 3

Let P1 be prime, when P1 is not equal to 3
P1 divided by 3 leaves 1 or 2
Then P2 is divided by 3 and the remaining 2 or 0, 0 is a composite number, so the remaining 2
Then P3 is divided by 3 and 0,2 + 4 = 6, so it is a composite number and does not exist
So P1 can only be 3
If the inequality log2 (- x2 ax) > 0 has a solution on R, the value set of a is a. if the function f (x) = ax2-2x + 2 has a zero point on [- 1,1], the value set of a is B
(1) Finding sets a and B
(2) Whether there is a value of a such that the inequality log2 (- x2 ax) ≤ 0 holds on X ∈ a ∩ B. if there is, the value of a does not exist
1)
From log2 (- x ^ 2-ax) > 0
The result is - x ^ 2-ax > 1
That is x ^ 2 + ax + 10, a ^ 2-4 > 0, a > 2 or A2 or X
Let P1, P2, P3 be three prime numbers, and P2 = P1 + 4, P3 = P1 + 8. Prove that P1 = 3
Let P1 be prime, when P1 is not equal to 3
P1 divided by 3 leaves 1 or 2
Then P2 is divided by 3 and the remaining 2 or 0, 0 is a composite number, so the remaining 2
Then P3 is divided by 3 and 0,2 + 4 = 6, so it is a composite number and does not exist
So P1 can only be 3
If the solution set of log2 ^ (AX ^ 2 + X + 3) > 0 is r, then the value range of a is r
log2(ax2+x+3)>0=log2(1)
When a = 0, the equation log2 (3) > 0 holds
When a ≠ 0, y = log2 (x) is an increasing function
So AX2 + X + 3 > 1
That is, AX2 + X + 2 > 0 holds on R
So the image can only be above the x-axis
a>0 ;△=1-8a1/8
In conclusion, the value of a is a = 0 or a > 1 / 8
A > 0 and ax ^ 2 + X + 3 > 1
If the solution set of ax ^ 2 + X + 2 > 0 is r, then the discriminant = 1-8a0
A kind of
Given the image of the first-order function y = x + m and the inverse scale function y = m + 1 / X (M is not equal to - 1), we can find the value of x0 at the intersection P (x0,3) of the first quadrant
Substitute x = x0, y = 3 into the above formula, where 3 = x0 + m, 3 = m + 1 / x0
The solution is x0 = 1
X0=1
If the inequality log2 (x-1) ^ 2
Zero
Best answer the answer to the prosecution's help
y=cos2x
The center of symmetry of cos2x is the point of intersection with the x-axis
cos2x=0
2x=kπ+π/2
x=kπ/2+π/4
Let k = - n
So it's (- n π / 2 + π / 4,0)
Now it's (- π / 2,0)
So a = [- π / 2 - (- n π / 2 + π / 4), 0]
That is, a = (n π / 2-3 π / 4,0)
It is known that the intersection point of the first quadrant of the image with the first-order function y equal to x plus m and the inverse scale function y equal to m / x plus 1 is p (x.3)
It is known that the intersection point of the first quadrant of the image where the first-order function y equals x + m and the inverse scale function y equals M / x + 1 is p (x.3). The analytic expressions of the first-order function and the inverse scale function are obtained
Two functions are y = x + m, y = (M + 1) / X. let P (x ', 3) be the intersection point, then x' + M = 3, (M + 1) / X '= 3,
From the second formula, we know that M = 3x '- 1, substituting into the first formula, X' = 1, M = 2
The expressions of the two functions are y = x + 2, y = 3 / X
Is the inverse scale function y = (M + 1) / x? Or y = m / x + 1
The idea is the same,
If y = m / x + 1
x+m=3 （1）
m/x+1=3 （2）
From (1) we can get x = 3-m, and from (2) we can get m / (3-m) + 1 = 3, and simplify 3M = 6, M = 2
So y = x + 2, y = 2 / x + 1
From the intersection point of two images P (x, 3), we can get (x > 0), 3 = x + m; 3 = (M + 1) / x, that is, x = 3-m, 3x = m + 1, we can get the solution, x = 1, M = 2. Therefore, the two analytic expressions are y = x + 2 and y = 3 / X respectively
Simultaneous y = - x + 8, y = 12 / X
-x+8=12/x
x^2-8x+12=0
(x-2)(x-6)=0
x1=2,x2=6
So: Y1 = 6, y2 = 2
A(2,6),B(6,2)
The midpoint of AB is C (4,4)
Obviously, Ao = Bo, the triangle AOB is isosceles triangle, so: OC is perpendicular to ab
OC = (4 ^ 2 + 4 ^ 2) ^ (1 / 2) = 4 (radical 2)
A. unfold
Simultaneous y = - x + 8, y = 12 / X
-x+8=12/x
x^2-8x+12=0
(x-2)(x-6)=0
x1=2,x2=6
So: Y1 = 6, y2 = 2
A(2,6),B(6,2)
The midpoint of AB is C (4,4)
Obviously, Ao = Bo, the triangle AOB is isosceles triangle, so: OC is perpendicular to ab
OC = (4 ^ 2 + 4 ^ 2) ^ (1 / 2) = 4 (radical 2)
AB = ((2-6) ^ 2 + (6-2) ^ 2) ^ (1 / 2) = 4 (radical 2)
Area of triangle OAB = (1 / 2) AB * OC = (1 / 2) * 4 * 4 * 2 = 16
Y is equal to 2x divided by 3x minus 4 range how to find, urgent!
Such as the title
Because y = 2x / (3x-4)
(3x-4)y=2x
x=4y/(3y-2)
So the range: y belongs to R and Y is not equal to 2 / 3
What is the X domain???
In every quadrant, the intersection of the linear function y = x + m and the inverse scale function y = m + 1 / X (M is not equal to - 1) is p (Z, 3)
Bring P (Z, 3) into
Linear function 3 = Z + m (1)
Inverse scale function 3 = m + 1 / Z (2)
(1)-(2)
3-3=z+m-m-1/z
z=1/z
z^2=1
z=1 or z=-1
Z = 1 into (1) 3 = 1 + m m = 2
Z = - 1 to (1) 3 = - 1 + M = 4
Do you know how to send a message to me
Y is equal to 3x square minus 5x plus 2, X belongs to bracket 2,6, find the range of function (expressed by interval)
y=3(x-5/6)²-1/12
Two
It's broken down
4x-7=0，5x+7=0
x=7/4，x=-7/5
Two