Let the definition domain of odd function f (x) be r, and the minimum positive period T = 3. If f (1) ≥ 1, f (2) = 2A − 3A + 1, then the value range of a is () A. A ＜ 1 or a ≥ 23B. A ＜ - 1C. − 1 ＜ a ≤ 23d. A ≤ 23

Let the definition domain of odd function f (x) be r, and the minimum positive period T = 3. If f (1) ≥ 1, f (2) = 2A − 3A + 1, then the value range of a is () A. A ＜ 1 or a ≥ 23B. A ＜ - 1C. − 1 ＜ a ≤ 23d. A ≤ 23

If f (1) ≥ 1, f (2) = 2A − 3A + 1, f (2) = f (- 1) ≤ - 1, 2a − 3A + 1, 2a − 3A + 1, 3a-2 ≤ 0, 3a-2 ≤ a ≤ 23, and a + 1 ≠ 0, 3a-1 < a ≤ 23, C is selected

If the absolute value of a is 7 and B is 2 larger than the opposite number of 7, then what is the sum of a + B?

This is the simplest math problem I've ever seen

The logarithmic function ㏑ (a + x) / (1-x) is an odd function, and the value range of a is obtained

f(x)=㏑(a+x)/(1-x)
f(-x)=ln(a-x)/(1+x)=-f(x)=-ln(a+x)/(1-x)=ln[(a+x)/(1-x)]^(-1)=ln(1-x)/(a+x)
So a = 1

+/4 / what is the reduced number? / / is the absolute value sign
+/-4/

4

If the function f (x) defined in the interval (- 1,1) satisfies 2F (x) - f (- x) = LG (x + 1), then f (x)=______ ．

∵ 2F (x) - f (- x) = LG (x + 1), ① ∵ 2F (- x) - f (x) = LG (- x + 1), ② ① × 2 + ②, 3f (x) = 2lg (x + 1) + LG (1-x) ∵ f (x) = 23lg (x + 1) + 13lg (1 − x), so the answer is 23lg (x + 1) + 13lg (1 − x)

It is known that the square of 1 + X over F (x) = ax + B is an odd function defined on (- 1,1), and f (1 / 2) = 2 / 5
Finding the analytic expression of function f (x)
If the function f (x) is an increasing function in the interval [- 1, A-2], find the value range of A

1. The odd function of F (x) on (- 1,1)
∴f(0)=b=0
And ∵ f (1 / 2) = (A / 2 + b) / (1 + 1 / 4) = 2 / 5
∴a=1
∴f(x)=x/(1+x^2)
2. Set - 1

Given that the domain of F (x) is [0,1], find the domain of F (x) = f (2x) / (4x-1)

[0,2/3]

A mathematical problem of rational number
Expand (x ^ 2-x + 1) ^ 6 to get a12x ^ 12 + a11x ^ 11 + +A2x ^ 2 + a1x + A0, then A12 + a11 + +A2 + A1 + A0 =

Expand (x ^ 2-x + 1) ^ 6 to get a12x ^ 12 + a11x ^ 11 + +A2x ^ 2 + a1x + A0 is (x ^ 2-x + 1) ^ 6 = a12x ^ 12 + a11x ^ 11 + +A2x ^ 2 + a1x + A0, so if x is taken as 1 in the left formula, there will be: (1-1 + 1) ^ 6 = A12 + a11 + +a2+a1+a0a12+a11+… +a2+a1+a0=1...

Draw the points representing each logarithm on the number axis and compare their sizes:
(1) - 8, - 6; (2) - 5,0.1,; (3) - 1,0 of 4 parts;
(4) - 4.2, - 5.1; (5) 2 of 3,3 of 2; (6) + 1,0 of 5

No, it's very simple. Ask the people around you!

Given that x = 2 is the solution of the equation AX-2 = 3x + 1 about X, find the value of A

Take x = 2 into the equation AX-2 = 3x + 1,
Then 2a-2 = 7,
2a=9,
The solution is a = 9 / 2