We know that the system of inequalities x + 1 / X-1 ≥ 0 on X, ① (x-2a + 1) (x-a ^ 2) ≤ 0, ② where a ∈ R. (1) if the solution set of the system of inequalities is an empty set, find the range of A (2) If the solution set of inequality system is a nonempty set {x | B ≤ x ≤ - 1}, the value range of real number B is obtained

We know that the system of inequalities x + 1 / X-1 ≥ 0 on X, ① (x-2a + 1) (x-a ^ 2) ≤ 0, ② where a ∈ R. (1) if the solution set of the system of inequalities is an empty set, find the range of A (2) If the solution set of inequality system is a nonempty set {x | B ≤ x ≤ - 1}, the value range of real number B is obtained

The first solution of the inequality is x > 1 or X ≤ - 1; the second solution is 2a-1 ≤ x ≤ a ^ 2
Because the solution set is empty: so a ^ 2 ≤ 1 and 2a-1 > - 1, the solution set of 01 is 2a-1 ≤ x ≤ - 1 or 1

Solving inequality, x ^ 2 - (2 + a) x + 2A > 0

(x-2)(x-a)>0
1) When a > 2, the solution is {x | x > A, X

Finding the solution set of inequality 3 (x-1) > ax

3(x-1)>ax
3x-3>ax
(3-a)x>3
When a = 3, the inequality has no solution
When A3 / (3-A)
When a > 3, X

The inequality (AX-1) (x + 1) / (X-2) (x-3) > 0
Help, hurry!

(ax-1)(x+1)/(x-2)(x-3)>0
That is, (AX-1) (x + 1) * (X-2) (x-3) > 0
When a = 0, (x + 1) (X-2) (x-3)

Given the function f (x) = x + 9 / X (1), judge the monotonicity of F (x) on (0, positive infinity) and prove it. (2) find the domain value of F (x)

Solution;
(1) The derivative function is right
So it decreases monotonically on (0,3) and increases monotonically on (3, positive ∞)
2. Domain {X / X ∈ R, X ≠ 0}

Given that the definition field of function f (x) = x / (1 + x ^ 2) is (- 1,1), its monotonicity is proved

Definition method:
Let - 10, the function increases monotonically in the interval (- 1,1)

∫dy/ylny=∫dx/x
ln|lny|=ln|x|+lnc
The solution of ∫ DX / X is not ln | x | + C
Why ln|x| + LNC

Sequence 1 / 1 * 2 + 1 / 2 * 3 + +1/n(n+1)
Sn = 1-1 / 2 + 1 / 2-1 / 3 + --- + 1 / n-1 / (n + 1) = 1-1 / (n + 1)
How can I get the 1 - in 1-1 / (n + 1)? Is n of 1 / N - 1,
You don't have to understand the complexity:
Sn=(1-1/2)+(1/2-1/3)+...+(1/n-1-1/n)+(1/n-1/n+1)=1+(-1/2+1/2)+(-1/3+1/3)+...+(-1/n+1/n)-1/n+1
=1-1/(n+1)

Let y = arcsinx + lntanx, find dy / DX

dy/dx=1/√(1+x^2)+sec^2x/tanx

The monotone decreasing interval of y = x-arcsinx is

There are two methods. First, if you are familiar with the derivative, you can directly seek the derivative, which involves the derivative of arcsinx. Its derivative is 1 divided by the root sign 1-x square. If you don't know, you can find the derivative. Because y = arcsinx, so x = siny, for the derivative of X, DX / dy = cosy, that is, dy / DX = 1 / cosy

Let the following function be reduced to the form of y = a (X-H) ^ 2 + K (1) y = 3x ^ 2 + 12x-3 (2) y = 2
Change the following function into y = a (X-H) ^ 2 + K
(1) Y = one 3x ^ 2 ten 12x-3
(2)y=2x ^2+8x-6
(3)y=2/1x^2-2x-1

(1) Y = one 3x ^ 2 ten 12x-3
=-3（x^2-4x）-3
=-3(x^2-4x+4-4)-3
=-3(x-2)^2+9
(2)y=2x ^2+8x-6
=2(x^2+4x)-6
=2(x^2+4x+4-4)-6
=2(x+2)^2-14
(3)y=2/1x^2-2x-1
=2/1 (x^2-4x)-1
=2/1 (x^2-4x+4-4)-1
=2/1(x-2)^2-3