If the square of root 2a-8 plus the absolute value of B-1 = 0, find the values of a and B

If the square of root 2a-8 plus the absolute value of B-1 = 0, find the values of a and B

The absolute value of square-8 of root 2A plus B-1 = 0,
because
If the root is greater than or equal to 0, the absolute value is greater than or equal to 0, and to be equal to 0, only
2a^2-8=0,b-1=0
therefore
A = 2 or - 2, B = 1

If the real numbers a and B satisfy the absolute value of the square of 2A + B + the square of B-10 = 0, then AB =?

2a+b^2=0
And B ^ 2-10 = 0 leads to B ^ 2 = 10, B = positive and negative root 5 2A + 10 = 0, so a = - 5
AB = root 5 of plus or minus 5 times
I don't know the question, right?

The nonzero real numbers a and B satisfy | 2A − 4 | + | B + 2|+ (a − 3) B2 + 4 = 2A, then a + B is equal to______ .

∵a≥3，
| the original equation can be changed into | B + 2|+
(a−3)b2＝0，
ν B + 2 = 0 and (A-3) B2 = 0,
∴a=3，b=-2，
∴a+b=1．
So the answer is 1

Let a, B, C be the square + absolute value of real number √ 2A + 1 + (B-3), C-2 = 0. Find the value of ABC / 3 √ 2A + 1 is the square of part (B-3) is a part of absolute value, C-2 is a part. Don't make these three formulas under the root sign!

Because √ (2a + 1) + (B-3) ^ 2 + | C-2 | = 0,
Because √ (2a + 1) ≥ 0, (B-3) ^ 2 ≥ 0, | C-2 | 0,
So √ (2a + 1) = 0, (B-3) ^ 2 = 0, | C-2 | = 0,
So 2A + 1 = 0, B-3 = 0, C-2 = 0,
So a = - 1 / 2, B = 3, C = 2,
So the third root sign (ABC / 3) = cubic root - 1 = - 1

Given that the real number a satisfies a + 2 the square of a + the cubic root of a = 0, find the absolute value of A-1 + the absolute value of a + 2A

Let's know that the real number a satisfies the square of a + 2 + the cubic root of a = 0, 3 √ a 3 = a;  2A + 2 √ a  a  a  a ﹤ a ﹤ a ﹤ a ﹤ a ≤ 0; find the absolute value of A-1 + the absolute value of a + the value of 2A = 1-a-a + 2A = 1

It is known that a, B are real numbers, and the absolute value of root 2A + 6, b-radical 2 = 0, the process of solving X: (a + 2) x + b square = A-1 requires proving the process!

According to the meaning of the question: 2A + 6 = 0, b-radical 2 = 0 (because the value of the root and absolute value is non negative, the two nonnegative addition = 0, indicating that both parts = 0), so a = - 3, B = root 2, bring a = - 3, B = root 2 into the equation to get: (- 3 + 2) x + the square of root 2 = - 3-1, and the solution is: x = 6
Please accept if you are satisfied

If the square of (a + Radix 2) and / B-1 / are opposite to each other, then the value of 1 / B-A is?

Opposite numbers add up to 0
If one is greater than 0, the other is less than 0
So both are equal to zero
So a + √ 2 = 0, B-1 = 0
a=-√2,b=1
1/(b-a)
=1/(√2+1)
=√2-1

If the square of (a + 2) is opposite to the absolute value of B-1, what is the value of B-A? All right, score

Because the absolute value of sum of squares is positive, it can only be 0. If a = - 2, B = 1, then B-A = 3

If the absolute value of B-1 and the square of [(radical 3) - A] are opposite to each other, then the value of a + B is

The absolute value of B-1 is opposite to the square of [(radical 3) - A]
therefore
b-1=0
√3-a=0
Namely
a=√3
B=1
therefore
a+b=√3+1

Let a B be a real number and satisfy the following conditions: root sign (1 + a) - (B-1) multiply root sign (1-B) = 0 to find the value of a ^ 2006-b ^ 2007 Let a B be a real number and satisfy the following conditions: (root (1 + a)) - (B-1) multiply (root (1-B)) = 0 to find the value of a ^ 2006-b ^ 2007 Come on, come on. We need to answer in detail

√（1+a）-(b-1)*√(1-b)=0
√(1+a)+√[(1-b)^3]=0
Root sign constant > = 0
To make the equation true, we can only
1+a=0
(1-b)^3=0
a=-1,b=1
therefore
a^2006-b^2007
=(-1)^2006-1^2007
=0