Given that a, B, C satisfy 1 / 2 times the absolute value of A-B + the square - C + 1 / 4 = 0 of the root sign 2B + C + C, find the square root of - A (B + C)

Given that a, B, C satisfy 1 / 2 times the absolute value of A-B + the square - C + 1 / 4 = 0 of the root sign 2B + C + C, find the square root of - A (B + C)

I think the known condition is: (1 / 2) ︱ A-B ︱ + √ (2B + C) + C ^ 2-C + 1 / 4 = 0
If so, the method is as follows:
∵（1/2）｜a－b｜＋√（2b＋c）＋c^2－c＋1/4＝（1/2）｜a－b｜＋√（2b＋c）＋（c－1/2）^2＝0,
A-B = 0, 2b + C = 0, C-1 / 2 = 0 must be established at the same time
From C-1 / 2 = 0, the following results are obtained: C = 1 / 2, ν 2B + 1 / 2 = 0, νb = - 1 / 4, ν a - (- 1 / 4) = 0,  a = - 1 / 4
The square root of - A (B + C) = - (- 1 / 4) (- 1 / 4 + 1 / 2) = 1 / 16, and the square root of - A (B + C) is 1 / 4 or - 1 / 4
Note: if the situation is not what I guess, please add

Given the absolute value A-B + 1 + root sign 3a-2b-1 = 0, find the arithmetic square root of 4A + 5B to the second power

∵ absolute value A-B + 1 + radical 3a-2b-1 = 0,
∴a-b+1=0
3a-2b-1=0
∴a=3
B=4
The arithmetic square root of the second power of 4A + 5B = √ (4a + 5b) 2 = √ (12 + 20) mm2 = 32

Given the absolute value of the square root of a + B + the third power of B-8) = 0, find the square root of - 2A of B and the cube root of 4A of B

√(a+b)^2+|b^3-8|=0,
A + B = 0, B ^ 3-8 = 0,
The solution is: a = - 2, B = 2,
ν b-2a = 6, and its square root is ±√ 6,
4A / b = - 4, and its cube root is - 3 √ 4

Given that a and B satisfy the absolute value of the radical A-2 + (a + 2b) - 4, find the arithmetic square root of B2 + 4a

You don't seem to understand this question. The format is wrong. I can't understand it

Given that the square root of 2a-1 is plus or minus 3, the cube root of 3A + B-9 is 2, and C is the integral part of root 57. Find the arithmetic square root of a + 2B + C

Because the square root of 2a-1 is plus or minus 3, 2a-1 = 9, so a = 5
Because the cube root of 3A + B-9 is 2, so 3A + B-9 = 8, so B = 2
Because 49 ＜ 57 ＜ 64, so 7 ＜ radical 57 ＜ 8, so C = 7
So a + 2B + C = 5 + 2 * 2 + 7 = 16
So the arithmetic square root of a + 2B + C is ± 4

It is known that a = a + 1 root sign a is the arithmetic square root of a, B = 4A + B times root sign B + 1 is the cube root of B + 1, find the square root and cube root of a + B

Because 1-B degree √ A is the arithmetic square root of a, so 1-B is equal to 2. Because 4A + B degree √ B + 1 is the cube root of B + 1, 4A + B is equal to 3. The equation system can be obtained as follows: 1-B = 24a + B = 3, a = 1, B = - 1, so a = √ 1 = 1, B = 3 √ (- 1 + 1) = 3 √ 0 = 0A + B = 1 + 0 = 1 because (± 1) 2 = 1, 1

If a = A-1, the arithmetic square root of the root sign 2B + 5 is the arithmetic square root of 2B + 5, and B = - 3B + 9, and the root sign 1-3a is the cube root of 1-3a, calculate the value of a-b

From the question: A-1 = 2 a = 3
-3b+9=3 b=2
A = radical 2B + 5 = 3
Cube root of B = 1-3a = - 2
A-B=3-(-2)=5

It is known that the square root of 2a-1 is the positive and negative root sign 3, and the square root of 3a-2b + 1 is plus or minus 3. Find the square root of 4a-b

2a-1=3
A=2
3a-2b+1=7-2b=9
b=-1
4a-b=9

Given that a = - 2-1 under the root sign, B = 2-1 under the root sign, find the value of degree 3 A, B + 3; know that the square root of 2a-1 is positive and negative 3, and the square root of 3a-2b is positive and negative 3, find 4a-b

The value of 3 times a, B + 3 times is not understood
The second question is OK
The meaning of the title is obtained;
2a-1=9
3a-2b=9
∴a=5 b=3
∴4a-b=20-3=17

If (radical A-4) + absolute value B-9 = 0. Find the square root of B of A

The root sign A-4 is a non negative number, and the absolute value B-9 is also a non negative number. The sum of two non negative numbers is 0, which means that both non negative numbers are equal to 0, so A-4 = 0, B-9 = 0, so a = 4, B = 9, so the square root of B in a = plus or minus 3 / 2