To solve the inequality ax ^ 2-2x + 1 > 0 about X, we should discuss a and what range should a take

To solve the inequality ax ^ 2-2x + 1 > 0 about X, we should discuss a and what range should a take

A:
ax^2-2x+1>0
Discriminant = (- 2) ^ 2-4a = 4 (1-A)
1) When a = 0: - 2x + 1 > 0, X0
The equation AX ^ 2-2x + 1 = 0 always has two unequal real roots
x=[2±2√(1-a)]/(2a)
=[1±√(1-a)]/a
The inequality is reduced to: - ax ^ 2 + 2x-1

In the interval [- 1,1], take any two real numbers a and B, and the equation x ^ 2 + ax-b ^ 2 = 0.1, find the probability 2 that the equation has real roots, and the probability 2 that the equation has two positive real roots
Take any two real numbers a and B in the interval [- 1,1], and the equation x ^ 2 + ax-b ^ 2 = 0
1. Find the probability that the equation has real roots
2. The probability that the equation has two positive real roots
In the interval [- 1,1], take any two real number a equations x ^ 2 + ax + B ^ 2 = 0
1. Find the probability that the equation has real roots
The probability of equation having two positive real roots

1, a ^ 2-4 * B ^ 2 > = 0 | a | > = 2 | B | draw the coordinate system p = 1 / 4
2, the same p = 1 / 16
If you don't know, ask again

If the real numbers x, y satisfy (x square + y square + 2) (x square + y square - 1) = 0, then x square + y square =? Find the speed
It's 21:10

Test point: solving quadratic equation with one variable by substitution method
Analysis: let x 2 + y 2 = m, simplify the equation and get the value of M
Let x2 + y2 = m, the equation be (M + 2) (m-1) = 0 ∫ M1 = - 2, M2 = 1 ∫ x2 + Y2 ≥ 0 ∫ M1 = - 2, i.e. x2 + y2 = 1
Comments: This paper mainly examines the substitution method, that is, to take a formula as a whole, replace it with a letter, and implement equal substitution. Note that x2 + Y2 is a non negative number

What is the negation of the proposition "for any x to be a real number, the square of X is greater than zero"?

Inverse: there exists x belonging to R, so that the square of X is greater than 0
No: any x belongs to R, so that the square of X is not greater than 0
Inverse No: there is x belonging to R, so that the square of X is less than or equal to 0

In the interval [0,1], take three real numbers x, y, Z, event a = {x ^ 2 + y ^ 2 + Z ^ 2}

Use three-dimensional coordinate system to solve {(x, y, z) | x ^ 2 + y ^ 2 + Z ^ 2

If you take any two real numbers in the interval (0, 1), the probability that the sum of the two real numbers is greater than 13 is ()
A. 1718B. 79C. 29D. 118

If any two real numbers in the interval (0, 1) are counted as (x, y), the plane area corresponding to the point is the square shown in the figure below, in which the sum of the two real numbers is greater than 13, that is, the plane area with x + Y > 13 is shown in the shaded part in the figure below: the square area s = 1, the shaded part area s, the shaded part area s, the shaded part s, the shaded part s = 1-12 · 13 · 13 = 1718, the probability that the sum of the two real numbers is greater than 13, P = s, the shaded part s = 1718, so select a

If the real numbers x and y satisfy the condition x − y + 1 ≥ 0y + 1 ≥ 0x + y + 1 ≤ 0, then the maximum value of 4x · (12) y is ()
A. 2B. 1C. 12D. 14

Make the plane region represented by the inequality system, as shown in the figure, ∵ 4x · (12) y = 22x-y, let z = 2x-y, then y = 2x-z, - Z is the section of the straight line on the Y axis. The larger the intercept, the smaller the Z. according to the graph, when the objective function passes through point B, the maximum value of Z can be B (0, - 1) from y + 1 = 0x + y + 1 = 0. At this time, z = 1, so the maximum value of 4x · (12) y = 22x-y is 2, so select a

Line L: ax-y-i = 0 and hyperbola C: x ^ 2-2y ^ 2 = 1 intersect at two points of PQ, whether there is a real number a, so that the circle with PQ as the diameter crosses the origin! Explain the reason
There is evaluation, there is no reason?

The line L: ax-y-1 = 0 intersects the hyperbola C: x ^ 2-2y ^ 2 = 1 at P and Q. when a is the value, the square of PQ distance = 4 + 4A ^ 2?

Substituting y = AX-1 into hyperbolic equation, sorting out, (1-2a ^ 2) x ^ 2 + 4ax-3 = 0 according to Weida's theorem, X1 + x2 = 4A / (2a ^ 2-1), x1x2 = 3 / (2a ^ 2-1) y1-y2 = a (x1-x2) the square of PQ distance = (x1-x2) ^ 2 + (y1-y2) ^ 2 = (1 + A ^ 2) (x1-x2) ^ 2 = 4 + 4A ^ 2 = 4 (1 + A ^ 2), so 4 = (x1-x2) ^ 2

It is known that the circle C: (x-1) square + (Y-2) square = 25 and the straight line L: (3m + 2) x + (M + 1) y = 10m + 7 (M belongs to R)
It is known that circle C: (x-1) square + (Y-2) square = 25 and straight line L: (3m + 2) x + (M + 1) y = 10m + 7 (M belongs to R) (1) it is proved that no matter what real number m takes, straight line L and circle C always intersect; (2) the equation of straight line when l is cut by circle C is obtained

Straight line L: (3m + 2) x + (M + 1) y = 10m + 7 constant crossing point Q (3,1)
The center of circle C: (x-1) & #178; + (Y-2) & #178; = 25 is C (1,2)
|QC | = root 5