The definition domain of the function f (x) = the square of 1-x under the root sign + the square of X under the root sign is

The definition domain of the function f (x) = the square of 1-x under the root sign + the square of X under the root sign is

If the function is meaningful, then 1-x ^ 2 > = 0, x ^ 2-1 > = 0
The solution is x = 1
The domain is defined as {1}

The function f (x) = the square of X - 3x + 2 is defined as

Is it under root sign (x ^ 2-3x + 2)
That is, find x ^ 2-3x + 2 > = 0
That is (x-1) (X-2) > = 0
That is, x > = 2 or X

F (x under root) = x + 1, f (x) =? Function f (x) = 1-x square under root + X-1 under root sign

f(√x)=x+1
Let t = √ x note that t ≥ 0
Then x = t
f(t)=t²+1 (t≥0)
therefore
f(x)=x²+1（x≥0)
1-x²≥0
x²-1≥0
So, x? - 1 = 0, x = 1 or - 1
The definition field is {x ｜ x = 1 or x = - 1}

When f (x) = radical (x + 6 / 1) - 1 is a set, and the domain of G (x) = LG (- x ^ 2 + 2x + m) is set M = 3, the intersection of complement set a and B is obtained

F (x) = under the radical sign {(5-x) / (x + 1)} {(5-x) / (x + 1)} ＞ = 0 ᙽ 5-x ＞ = 0 x + 1 ＞ 0 (∵ x + 1 in denominator, ≠ 0) x ＜ = 5 x ＞ - 1

The function f (x) = 3x2 The domain of 1 − x + LG (3x + 1) is______ .

For the function f (x) = 3x2
1−x+lg(3x+1)
The independent variable x needs to satisfy 1-x > 0 and 3x + 1 > 0, that is − 1
3＜x＜1，
So the answer is (− 1)
3，1)．

Find the definition domain of the following function (1) f (x) = 4-2x square (2) f (x) = LG (X-2) of x-3

One
4-2x^2>=0
The solution is: root 22 and X is not = 3
The domain is (2,3) U (3, + infinity)

Given the complete set u = R, function y= x-2+ The definition domain of X + 1 is a and function y= 2x+4 The definition domain of x-3 is B (1) Find the set a, B （2）（∁UA）∪（∁UB）．

(1) By
x-2≥0
x+1≥0 ⇒x≥2　　
A={x|x≥2}
from
2x+4≥0
X-3 ≠ 0} x ≥ - 2 and X ≠ 3
B = {x | x ≥ - 2 and X ≠ 3}
(2) A ∩ B = {x | x ≥ 2 and X ≠ 3}
∪ (cub) = Cu (a ∩ b) = {x | x ＜ 2 or x = 3}

Given the complete set u = R, let the definition domain of function y = LG (x-1) be set a and function y= If the value range of x2 + 2x + 5 is set B, then a ∩ (∁∪ b) = () A. [1，2） B. [1，2] C. （1，2] D. （1，2）

From the meaning of the title, X-1 > 0, x > 1,
∴A=（1，+∞），
From x2 + 2x + 5 ≥ 4
x2+2x+5≥2，
∴B=[2，+∞），∴∁∪B=（-∞，2）
∴A∩（∁∪B）=（1，2）．
Therefore, D

The definition domain of the complete set u = R, the function f (x) = 1 / radical (x + 2) + LG (3-x) is set a Set B = {x| x ^ 2-A

A:x+2>0
And 3-x > 0
The solution is: - 2

Inequality f (x)= 2+x The definition domain of X − 1 is set a, and the inequality about X (1 2) The solution set of 2x ＞ 2-a-x, (a ∈ R) is B. the value range of real number a such that a ∩ B = B is obtained

By 2 + X
X − 1 ≥ 0 leads to X ≤ - 2 or x > 1
So a = (- ∞, - 2] ∪ (1, + ∞)
(1
2)2x＞2−a−x⇔(1
2)2x＞(1
2)a+x⇔2x＜a+x⇔x＜a．
So B = (- ∞, a)
Because a ∩ B = B,
So B ⊆ a,
So a ≤ - 2, that is, the value range of a is (- ∞, - 2]