Let a belong to R, and solve the inequality about X (a quadratic - 1) x quadratic - 2aX + 1 > 0 Urgent need

Let a belong to R, and solve the inequality about X (a quadratic - 1) x quadratic - 2aX + 1 > 0 Urgent need

A:
(a^2-1)x^2-2ax+1>0
[(a-1)x-1]*[(a+1)x-1]>0
x1=1/(a-1)
x2=1/(a+1)
Because: A-10, x > - 1 / 2
2) When a = 1:
-2x+1>0,x
The inequality is written as: (a ^ 2-1) * x ^ 2 - 2aX + 1 & gt; 0A belongs to R. the solution set of the inequality is related to A. A ^ 2-1 = 0 is divided into several value ranges. It is a first-order inequality a ^ 2-1 & gt; 0. If the discriminant is less than or equal to 0, there is no solution. If the discriminant is greater than 0, there is a solution a ^ 2-1 & gt; 0. For example, 2 & nbsp; the final result has five value ranges of a, corresponding to five solution sets
If a = 1
The original inequality is equivalent to - 2x + 1 > 0, the solution is x0, the solution is x > - 1 / 2
If a ≠ ± 1
The original inequality is equivalent to (AX-1) ^ 2-x ^ 2 > 0
That is [(a + 1) X-1] [(A-1) X-1] > 0
When A-1 / 2
If a ≠ ± 1
The original inequality is equivalent to (AX-1) ^ 2-x ^ 2 > 0
That is [(a + 1) X-1] [(A-1) X-1] > 0
When a
The solution of the inequality about X: two thirds of x-three thirds of X is greater than or equal to 1 (a ≠ 0)
If the denominator is removed, 4ax-9 ≥ 6
4 ax ≥ 15 (2)
If the coefficient is reduced to 1, X ≥ 15 / 4A
Question: (1) in the process of solving the above problems, the basis of step 1 is:
(2) In the process of solving the above problem, which step did you start to make mistakes? Please write down the code of this step:
(3) The reason for the error is:
Please accept as a satisfactory answer
If the function f (x) = (A-1) x square + 2aX + A + 2 has an intersection with the X axis, then the value range of the real number a is?
When a = 1, f (x) = 2x + 3 has an intersection with the X axis
When a is not equal to 1
（2a)^2-4(a-1)(a+2)≥0
The solution is a ≤ 2
△≥0
4a²-4x1x（a+2）≥0
A ≤ - 1 or a ≥ 2
The range of F (x) = (3x-1) / (x + 3)
X + 3 is not equal to 0, so x is not equal to 3
When x = 3, f (x) = 4 / 3
So the range is (negative infinity, 4 / 3) and (4 / 3, positive infinity)
X is not equal to 3 and X belongs to R
f（x)=3-10/（x+3）
And 10 / (x + 3) belongs to R and is not equal to 0
So it is the upper solution
FX should be (3x-1) divided by (x + 3) equal to 3x-1 divided by x plus 3x-1 / 3 equal to 1 + X. in other words, the region should be non negative
The range is - 25 / 3 to positive infinity
-10 / 3 to positive infinity
Given the square of function f (x) = - SiNx + SiNx + A. (1) when Dang f (x) = 0 has a real solution, find the value range of A
(2) If x belongs to R, 1 is less than or equal to f (x) is less than or equal to 17 / 4, the value range of a is obtained.
formula
f(x)=-(sinx-1/2)²+1/4+a
-1
Find the range of F (x) = x ^ 4-3x ^ 2-4, X ∈ [1.3]
f(x)=x^4-3x^2-4=（x-3/2)^2-25/4
When x = 3 / 2, the minimum value is - 25 / 4
When x = 3, there is a maximum of - 4
So the range is [- 25 / 4, - 4]
We use the maximum and minimum to find the solution. F '(x) = 4x ^ 3-6x = 0, find out x = 0, 2 / 2 root sign 6, negative 2 / 2 root sign 6 (rounding off). Find out f (x) when x = 0 and 2 / 2 root sign 6, then find out f (x) when x = 1,3, and pick out the maximum and minimum value
The range of function f (x) = (sin2xcosx) / (1-sinx)#
Answer [- 1 / 2,4)
f(x)=sin2xcosx/(1-sinx)=2sinx*(cosx*cosx)/(1-sinx)
=2sinx*(1-sinx)*(1+sinx)/(1-sinx)
=Sin2x is not equal to (sin2x)
=2*(sinx+sinx*sinx)
=2*[(sinx+1/2)^2-1/4]
Because - 1
The range of function y = 3xx2 + X + 1 (x < 0) is______ .
Therefore, the answer is only if [− 1 + 1] = (− 1 + 1) + 1
If the range of function y = LG (x2 + 2x + m) is r, then the range of M is. M ≤ 1
The range is r
Then true numbers go to all positive numbers
Then the minimum value of parabola should be less than or equal to 0
Otherwise, the positive number between 0 and the minimum value cannot be obtained
So the discriminant is greater than or equal to 0
4-4m≥0
m≤1
x^2+2x+m≥0
Δ=b^2-4ac≥0
4-4*1*m≥0
m≤1
Find the range: y = 3x / (X & # 178; + X + 1) (x < 0), the answer is y ∈ [- 3,0)
y=3x／(x²+x+1)
=3/(x+1+1/x)
∵x0,-1/x>0
The mean value theorem is as follows:
- x + (- 1 / x) ≥ 2 √ [(- x) * (- 1 / x)] = 2 (if and only if - x = - 1 / x, i.e. x = - 1, take the equal sign)
∴x1/(x+1+1/x)≥-1
That is, 0 > 3 / (x + 1 + 1 / x) ≥ - 3
The range of the function y = 3x / (X & # 178; + X + 1) is 0 > y ≥ - 3