1. If G (x) = f (x + a) + F (x-a), the domain of function g (x) is obtained 2. Find the domain of the function f (x) = root sign (x-2k) + 1 / root sign (X & # 178; - 4) I don't think I understand the function. I hope the great God can give me some advice or experience I will ask if I don't understand. I hope you can answer patiently~

1. If G (x) = f (x + a) + F (x-a), the domain of function g (x) is obtained 2. Find the domain of the function f (x) = root sign (x-2k) + 1 / root sign (X & # 178; - 4) I don't think I understand the function. I hope the great God can give me some advice or experience I will ask if I don't understand. I hope you can answer patiently~

(1) Because g (x) is composed of F (x + a) and f (x-a)
Then the domain of G (x) is determined by the domain of F (x + a) and f (x-a)
Because the definition field of F (x) is (0,1)
It shows that the value range of the independent variable is (0,1)
Note that the value range of the independent variable is independent of the form of the independent variable
Then f (x + a) has 0

Given that real numbers a, B and C satisfy a × a + B × B = 1, B × B + C × C = 2, C × C + a × a = 2, what is the minimum value of AB + BC + AC?
Please reach:
1. Please use the knowledge of grade one
two
3. If you want to use the formula, please write it in detail
I wish you all a happy new year and all the best. Congratulations on Facai!

First of all, simplify the writing, a ^ 2 = a * a, similarly, the formula used in B ^ 2 = b * B: A ^ 2 + B ^ 2 + C ^ 2 + 2Ab + 2Ac + 2BC = (a + B + C) ^ 2 because: A ^ 2 + B ^ 2 + C ^ 2 = (1 + 2 + 2) / 2 = 2.5, so (a + B + C) ^ 2 = a ^ 2 + B ^ 2 + C ^ 2 + 2Ab + 2Ac + 2BC = 2.5 + 2 * (AB + AC + BC) because (a + B + C) ^ 2 is greater than, etc

If the real numbers a, B and C satisfy a ^ 2 + B ^ 2 = 1, B ^ 2 + C ^ 2 = 1, a ^ 2 + C ^ 2 = 1, then the minimum value of AB + BC + AC is

a^2=b^2=c^2=1 /2
The minimum value of AB + BC + AC is - 1 / 2

It is known that AB is all negative real numbers, and 1 / A + 1 / B-1 / A-B = 0

Solution
1/a+1/b-1/(a-b)=0
1/a+1/b=1/(a-b)
(b+a)/ab=1/(a-b)
ab=(a+b)(a-b)=a²-b²
∴b/a
=ab/a²
=(a²-b²)/a²
=1-(b/a)²
∴(b/a)²+(b/a)-1=0
∴b/a=(-1+√5)/2
Or B / a = (- 1 - √ 5) / 2
∵ A.B. are all negative
The B / A is positive
∴b/a=(-1+√5)/2

Given that a and B are real numbers, and 1 / A + 1 / B-1 / (a-b) = 0, finding the value of B / A is as follows
I'll give it to you

1 / A + 1 / B-1 / (a-b) = 0 (first general division) (a + b) / AB - 1 / (a-b) = 0 shift (a + b) / AB = 1 / (a-b) cross multiplication (a + b) (a-b) = AB a ^ 2-B ^ 2 = AB (both sides divide by a ^ 2) 1-B ^ 2 / A ^ 2 = B / A

Given that real numbers a and B satisfy a + B + A, B = 4ab-1, the value of a and B can be obtained
Fast

4ab=2ab+2ab
So (A & sup2; - 2Ab + B & sup2;) + (A & sup2; B & sup2; - 2Ab + 1) = 0
(a-b)²+(ab-1)²=0
If the sum of the squares is 0, it is equal to 0
So A-B = 0, AB-1 = 0
a=b,ab=1
So a = 1, B = 1 or a = - 1, B = - 1

Given that real numbers a and B satisfy real numbers (a + b) ^ 2 + A + B-2 = 0, find the value of a + B

(a + b) ^ 2 + A + B-2 = 0 (a + B-1) (a + B + 2) = 0, a + B = 1 or a + B = - 2

Given that the real number AB satisfies | A-4 | + radical B + 2 = 0, find the value of the A-side of B

|A-4 | ≥ 0 radical, B + 2 ≥ 0 two parts can only be equal to 0, A-4 = 0, a = 4, B + 2 = 0, B = - 2, the a square of B = (- 2) ^ 4 = 16

Given that a and B are real numbers and satisfy a ^ 2 + 1 = 0, B ^ 2 + B + 1 = 0, find the value of (B / a) + (A / b)
It's urgent

A ^ 2 + A + 1 = 0, B ^ 2 + B + 1 = 0, a, B are two solutions of the equation x ^ 2 + X + 1 = 0, a + B = - 1, ab = 1 (B / a) + (A / b) = [a ^ 2 + B ^ 2] / AB = [(a + b) ^ 2-2ab] / AB, so the original formula = - 1

Let 3A = 4B = 36 and find the value of 2A + 1b

∵ 3A = 4B = 36, ∵ for 3A = 4B = 36, take the logarithm with the base of 6, and get 2A = log63, 1b = log62, ∵ 2A + 1b = log63 + log62 = log66 = 1