On the equation AX ^ 2-2 (a + 1) x + A-1 = 0 of X when the real number a satisfies________ When, Both equations are greater than 1

On the equation AX ^ 2-2 (a + 1) x + A-1 = 0 of X when the real number a satisfies________ When, Both equations are greater than 1

Let two be x1, X2, because both are greater than 1, so x1-1, x2-1 are greater than 0, so (x1-1) + (x2-1) > 0, and (x1-1) * (x2-1) > 0. According to the relationship between the root and the coefficient, there are X1 + x2 = 2 (a + 1) / a X1 * x2 = (A-1) / a solutions to the inequality (x1-1) + (x2-1) > 0x1-1 + x2-1 > 0x1 + x2-2
Is that the problem?
Ax & sup2; - 2 (a + 1) x + A-1 = 0 when a satisfies________ There are two unequal real roots, two equal real roots and no real roots?
Because the discriminant [-- 2 (a + 1)] ^ 2 -- 4a (a -- 1) = a ^ + A + 1 = (a + 1 / 2) ^ 2 + 3 / 4 is greater than or equal to 0
So when a is not equal to 0, the equation always has two real roots
Let these two be different: X1 and x2
Then X1 + x2 = 2 (a + 1) / a X1 * x2 = (a -- 1) / A
Because both equations are greater than 1
So X1... Unfolds
Because the discriminant [-- 2 (a + 1)] ^ 2 -- 4a (a -- 1) = a ^ + A + 1 = (a + 1 / 2) ^ 2 + 3 / 4 is greater than or equal to 0
So when a is not equal to 0, the equation always has two real roots
Let these two be different: X1 and x2
Then X1 + x2 = 2 (a + 1) / a X1 * x2 = (a -- 1) / A
Because both equations are greater than 1
So X1 -- 1 is greater than 0, X2 -- 1 is greater than 0
So (x1 -- 1) + (x2 -- 1) is greater than 0, (x1 -- 1) (x2 -- 1) is greater than 0
That is: X1 + x2 -- 2 is greater than 0, X1 * x2 -- (x1 + x2) + 1 is greater than 0
2 (a + 1) / a -- 2 is greater than 0 (1)
(A-1) / a -- 2 (a + 1) / A + 1 is greater than 0 (2)
From (1): A is greater than 0
From (2): A is less than 0
So there is no solution. Put it away
If the equation ax2-x-1 = 0 has a solution in (0,1), find the value range of real number a
(1) When a = 0, f (x) = - X-1, its zero point is - 1 ∉ [0, 1], 〈 a ≠ 0; & nbsp; & nbsp; (2) when a ≠ 0, the equation ax2-x-1 = 0 has exactly one solution in (0, 1), that is, the quadratic function f (x) = ax2-x-1 has exactly one zero point in (0, 1), and the solution is a > 2. Therefore, the value range of a is (2, + ∞)
In the interval [0,1], take any two numbers a and B, and the probability of two real numbers of equation x2 + ax + B2 = 0 is ()
A. 18B. 14C. 12D. 34
If the two parts of the equation x2 + ax + B2 = 0 are both real numbers, then: △ = a2-4b2 ≥ 0, that is: (a-2b) (a + 2b) ≥ 0, that is, the area of the region composed of a-2b ≥ 0 is 14, the area of the region composed of any two numbers a and B in the interval [0, 1] is 1, and the probability of the two parts of the equation x2 + ax + B2 = 0 being both real numbers is 14; therefore, B is selected
If the solution of ax square + BX + C ＜ 0 (a ≠ 0) is x ＜ 2 or X ＞ 3, find the solution of the inequality ax square - BX + C ＞ 0
The inequality is less than 0 and the solution set is x < 2 or x > 3
So a
What is the range of the square of the function y = x + 1 / 2 x
The range is y ≥ 2 because the function
y=x^2+1/(x^2)≥2{(x^2)[1/(x^2)]}^(1/2)=2
The minimum value is obtained when x = 1
This function is a lower bounded function and has no upper bound
[2, positive infinity]
What is the range of the function y = x & # 178; / (X & # 178; + 1)
Y (X & # 178; + 1) = x & # 178;, (Y-1) x & # 178; + y = 0, X & # 178; = - Y / (Y-1) ≥ 0, that is, Y / (Y-1) ≤ 0, so 0 ≤ y
The inequality ax square + BX + C about X is known
PS we know that the solution set of the inequality ax square + BX + C ＜ 0 about X is (x / X ＜ m or X ＞ n) (m ＜ n ＜ 0), then the solution set of the inequality CX square - BX + a ＞ 0 about X is?
cx²-bx+a>0
Divide both sides of the inequality by X & sup2 at the same time;
c-b/x+a/x²>0
a(-1/x)²+b(-1/x)+c>0
Because the solution of ax & sup2; + BX + C0 is m
1. The range of square + 1 of function y = x 2. The value of 1-x, x under the radical of function y = x + when it belongs to {- 3, - 2, - 1,0,1}
The range is Y & nbsp; ≥ 1
They are (- 1, √ & nbsp; 3-2, √ & nbsp; 2-1, 1, 1) respectively
The inequality ax-2ax + 2A + 3 about X is known
A is greater than 0, and the tower is less than 0
Find the range of the following functions: (1) y = - xsquare + X-1; (2) y = xsquare - 1 / xsquare + 1
The method of quadratic function collocation to find the range
y=-(x^2-x)-1=-(x-1/2)^2-3/4 ymax=-3/4 y
The 3-2x power of 2 is less than the 3x-4 power of 0.5