It is known that the solution set of inequality (a + 1) x > 2 is X

It is known that the solution set of inequality (a + 1) x > 2 is X

Divide both sides by a + 1
The solution set is X

The cube of (X-2) = -0.125
Find x
3 (x + 5) = - 375

1
(X-2)^3=-0.125=(-0.5)^3
X-2=-0.5
X=2-0.5=1.5
2
3(X+5)^3=-375
(X+5)^3=-125=(-5)^3
X+5=-5
X=-10

Given a (- 1,0), B (0,2), point P is any point on the circle C: (x-1) ^ 2 + y ^ 2 = 1, then the maximum area of triangle ABP
Detailed explanation

AB linear equation is: 2x-y + 2 = 0
Let a line parallel to ab: 2x-y + B = 0 be tangent to circle C
Then: (x-1) ^ 2 + (2x + b) ^ 2 = 1 has and has only one solution
5x^2+(4b-2)x+b^2=0
Discriminant △ = (4b-2) ^ 2-20b ^ 2 = - 4B ^ 2-16b + 4 = 0
b=-2±√5
Therefore, the line parallel to AB and tangent to circle C is 2x-y-2 ± √ 5 = 0
Among them, the farthest distance from AB is: 2x-y-2 - √ 5 = 0
The distance between them is: (4 √ 5 + 5) / 5
|AB|=√5
Therefore, the maximum area of triangle ABP is 1 / 2 * (4 √ 5 + 5) / 5 * √ 5 = 2 + √ 5 / 2

Please prove that: 111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111

Proof: if P = 5, obviously
If P ≠ 5, then (10, P) = 1
From Fermat's theorem,
10^p=10modp
10^p-1=9modp
Because (P, 9) = 1
So (10 ^ p-1) / 9 = 1modp
(10^p-1)/9-1=0modp
That is, the proposition is established

If the function f (x) = x2 + 2 (A-1) x + 2 is a decreasing function on (- ∞, 4], then the value range of real number a is ()
A. a≤-3B. a≥-3C. a≤5D. a≥5

∵ f (x) = x2 + 2 (A-1) x + 2 = (x + A-1) 2 + 2 - (A-1) 2, its axis of symmetry is: x = 1-A ∵ function f (x) = x2 + 2 (A-1) x + 2 is a decreasing function on (- ∞, 4], so a is selected

Given function f (x) = LG [x2 + (a + 1) x + 1]
(1) If the domain is r, find the range of real number a; (2) if the domain is r, find the range of real number a

(1) If the domain of definition is r, find the range of real number a;
For any real number x, let x2 + (a + 1) x + 1 > 0 be constant, delta = 0, a > = 1 or a

Fill in the brackets with the prime number: 70 = () times () times ()
Multiply means multiply sign

2*5*7

Let a two-dimensional random variable (x, y) have a joint probability density f (x, y) = {C (x + y) 0 ≤ y ≤ x ≤ 1, 0, others} then the constant C=

The solution is C = 2

If the function FX = x + BX + C has f (1-x) = f (1 + x) for any real number x, then the size relation of f1.f0.f3

∵f（1+x）=f（1-X）
The axis of symmetry of F (x) is x = 1
∴b=-2
And f (x) is a quadratic function with an opening upward
∴f(1)

Find out the greatest common factors 18 and 12 8 and 6 12 and 4 30 and 6 of the following numbers

They are 6, 2, 4, 6, 6, 2, 4, 6, 2, 4, 6, 2, 4, 6, 2, 4, 6, 2, 4, 6, 2, 4, 6, 2, 4, 4, 6, 2, 4, 6, 4, 4, 6, 4, 4, 4