Given the function f (x) = LG (3 ^ x + x-a), if x ∈ [2, + infinity], f (x) ≥ 0 is constant, then the value range of A

Given the function f (x) = LG (3 ^ x + x-a), if x ∈ [2, + infinity], f (x) ≥ 0 is constant, then the value range of A

(－∞,10]

Given the function f (x) = LG (2x-b) (B is a constant), if x ≥ 1, f (x) ≥ 0 is constant, then the value range of B is______ ．

∵ f (x) = LG (2x-b), when x ≥ 1, f (x) ≥ 0 is constant, and ∵ 2x-b ≥ 1 is constant for any x ∈ [1, + ∞), that is, B ≤ 2x-1, and when x ∈ [1, + ∞), t = 2x-1 is an increasing function, and the minimum value of T = 2x-1 is 1, so we can get B ≤ 1, that is, the value range of B is (- ∞, 1), so the answer is: (- ∞, 1]

Given the function f (x) = 2x3-3ax2 + A + B (where a and B are real constants) problem 1: discuss the monotone interval of function f (x);

f（x）＝2x3－3ax2＋a+b
f '（x）＝6x2－6ax=0
x=0 or x=a
Monotone interval of function f (x); (- ∞, 0) (0, a) (a, + ∞) (a > 0)
Monotone interval of function f (x); (- ∞, a) (a, 0) (0, + ∞) (a)

Function f (x) = LG (1-x2), set a = {x | y = f (x)}, B = {y | y = f (x)}, then the set represented by the shadow in the figure is ()
A. [-1，0]B. （-1，0）C. （-∞，-1）∪[0，1）D. （-∞，-1]∪（0，1）

The set a = {{x 124\\\\\\124\\\124\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\theregion of a ∪ B is a ∪ B. the set of the remaining elements after removing a ∩ B is: (- ∞, - 1] ∪ (0,1), so D

Given that in the function f (x) = LG (AX BX) + X, the constants A and B satisfy a > 1 > b > 0, and a = B + 1, then the solution set of F (x) > 1 is ()
It is written that the solution of ax BX ＞ 0, that is, (A / b) ^ x ＞ 1, leads to X ＞ 0, so the domain of definition of function f (x) is (0, + ∞),
How did this come out?
I can understand the following analysis, but I can't understand it
Because a ＞ 1 ＞ B ＞ 0, ax increases, and - BX increases, so t = ax BX increases,
And y = LGT is increasing, so f (x) = LG (AX BX) + X is increasing function,
And f (1) = LG (a-b) + 1 = LG1 + 1 = 1, so f (x) > 1 when x > 1,
So the solution set of F (x) > 1 is (1, + ∞)
So choose B

, a = B + 1, then AX = (B + 1) x, then LG (AX BX) + x = LG [(B + 1) x-bx] + x = LG (BX + x-bx) + x = lgx + X, lgx must have meaning, so x > 0 is like this, lgx is negative at (0,1), LG1 = 0, lgx is greater than 0 at (1, + ∞), LG1 + 1 = 1, ax BX > 0, get ax > BX, I can't get a / b.x

Function problem f (x) = LG (AX BX) (a > 1 > b > 0)
Finding the domain of F (x)

Because: (a-b) x > 0
And a > 1 > b > 0
So x > 0
So the definition field is: x > 0

Let I = R, f (x) = LG (x2-3x + 2) and G (x) = LG (x-1) + LG (X-2), then G ∪ CIF & nbsp; equals ()
A. （2，+∞）B. （-∞，2）C. [1，+∞）D. （1，2）∪（2，+∞）

The domain of ∵ f (x) = LG (x2-3x + 2) is f, the domain of function g (x) = LG (x-1) + LG (X-2) is g, ∵ f = {x | x2-3x + 2 ＞ 0} = {x | x ＞ 2, or X ＜ 1}, g = {x | x − 1 ＞ x − 2 ＞ 0} = {x | x ＞ 2}, ∪ CIF = {x | x ≥ 1}, so C

How to simplify the root sign, for example: √ 75 - √ 5 / 9 + √ 26

5 times the root number 1 / 5 + 1 / 2 times the root number 20 - 1 / 2 times the root number 1 / 5 + 45

5√(1/5)+1/2√20-5/2√(1/5)+√45
=√5+2√5-1/2√5+3√5
=(6-1/2)√5
=11/2√5

Third radical 3 + Half radical 2 - fifth radical 5

Original formula = 1.732/3 + 1.414/2-2.236/5 = 0.837