What is the definition domain of the 0 th power of the function y = ln (2x-1) + (x-1)?

What is the definition domain of the 0 th power of the function y = ln (2x-1) + (x-1)?

X is not equal to 1
2x-1 is greater than 0, X is greater than 1 / 2
So x is not equal to 1 and X is greater than 1 / 2

Find the domain of y = ln (x + 1); y = arcsin (LNX)

1) X + 1 > 0, so x > - 1
2) LNX = siny, so - 1=

Let y = SiN x / E to the power of X, and find dy

dy=[(cosx-sinx)/e^x]*dx

Y = ln root sign (1 + x square) + e's - x power to find dy

dy=y'dx=(x/(1+x^2)-e^(-x))dx

Let y = sin (the x power of 2x + e) be known to find dy

dy=cos(2x+e^2x)*(2+2e^2x) dx

Given the set a = {y | y = log2x, X ＞ 1}, B = {y | y = (1) 2）x，x＞1}， Find (1) a ∩ B; （2）（CRA）∪B．

(1) ∵ set a = {y | y = log2x, x > 1} = {y | Y > 0},
B={y|y=（1
2）x，x＞1}={y|0＜y＜1
2}，
∴A∩B={y|0＜y＜1
2}．
(2) It is known from (1) that:
CRA={y|y≤0}，B={y|y＜1
2}，
∴（CRA）∪B={y|y＜1
2}．

Set a = {(x, y) | y = a}, B = {(x, y) | y = BX + 1, b > 0, B ≠ 1}. If a ∩ B has only one true subset, then the value range of real number a is______ .

The set a = {(x, y) | y = a} is a straight line,
The set B = {(x, y) | y = BX + 1, b > 0, B ≠ 1} is a curve,
∩ set a ∩ B has only one true subset, ∩ a ∩ B has only one element, as shown in Fig
So the range of a is: a > 1
So the answer is: (1, + ∞)

If a = {x | x2-x-2 ≤ 0}, B = {x | y = ln (1-x)}, then a ∩ B = () A. （1，2） B. （1，2] C. [-1，1） D. （-1，1）

A={x|x2-x-2≤0}={x|-1≤x≤2}，
B={x|y=ln（1-x）}={x|1-x＞0}={x|x＜1}，
Then a ∩ B = {x | 1 ≤ x < 1} = [- 1, 1)
Therefore, C

The set a = {Y / y = (1 / 3) x power, X ∈ r} B = {Y / y = x 2, X ∈ r} Then a intersects B = ()

Y = (1 / 3) x power
Then a = {Y > 0}
B={y\y>=0}
A cross B = {Y / Y > 0}

0

To make y = Log1
2 (x + 3) (2 − x) is meaningful and requires (x + 3) (2-x) > 0
That is (x + 3) (X-2) < 0, the solution is - 3 < x < 2;
From EX-1 ≥ 1, X-1 ≥ 0, i.e. x ≥ 1
So a = {x | 3 ＜ x ＜ 2}; b = {x | x ≥ 1}
(1) A ∪ B = {x | 3 < x < 2} ∪ {x | x ≥ 1 = x | 3 ﹤ x < 2 or X ≥ 1} = {x | x ＞ - 3}
(2) ∵ CUA = {x | x ≤ - 3 or X ≥ 2}
∩ B = {x | x ≤ - 3 or X ≥ 2} ∩ {x | x ≥ 1 = x | x ≥ 2}