if If 2A − 2 and | B + 2 | are opposite numbers to each other, then (a-b) 2=______ .

if If 2A − 2 and | B + 2 | are opposite numbers to each other, then (a-b) 2=______ .

A kind of
2A − 2 and | B + 2 | are opposite numbers to each other,
Qi
2a−2+|b+2|=0，
∴2a-2=0，b+2=0，
A = 1, B = - 2,
∴（a-b）2=[1-（-2）]2=9．
So the answer is: 9

If the absolute values of root 2a-2 and B + 2 are opposite to each other, what is the square of a + B

From the meaning of the title
So the absolute value of root 2a-2 plus B + 2 is 0 (because they are opposite numbers to each other)
So 2a-2-0, B + 2 = 0
So a = 1. B = - 2
So a + B = - 1

If the absolute value of (a-b + 1) is opposite to a + 2A + 4 under the root sign, what is the square of (a + b)?

The absolute value of (a-b + 1) is opposite to a + 2A + 4 under the radical sign
|A-B + 1 | + √ (a + 2B + 4) = 0 changes 2a to B
a-b+1=0
a+2b+4=0
∴a=-5/3
b=-2/3
(a+b)²=(-5/3-2/3)²=49/9

What is the absolute value of radical 6 - radical 7

Radical 7-root 6

A mathematical problem. There is a p (7 / 6), only the answer, the root sign x2 + 4 is equal to the absolute value x + 3 In the rectangular coordinate plane, O is the coordinate origin, the image of the quadratic function y = - x2 + (k-1) x + 4 intersects point a with y axis, and points B with negative half axis of X axis, and s △ OAB = 6 (1) Find the coordinates of point a and point B; (2) Find the analytic expression of the quadratic function; (3) If point P is on the x-axis and △ ABP is an isosceles triangle, find the coordinates of point P

(1) From the function expression, C = 4, the intersection point a of quadratic function image and Y axis is (0,4), OA = 4
A is on the y-axis, B is on the x-axis, so OA ⊥ ob
S△OAB=OA×OB/2=2OB=6
So ob = 3
Because B is on the negative half axis of the X axis, B (- 3,0)
(2) Replace the coordinates of point B:
-（-3）²-3（k-1)+4=0
-3k=2
k=-2/3
y=-x²-5x/3+4
(3) P on the X axis, let P (x, 0)
AP²=X²+4²=X²+16
BP²=（X+3）²=X²+6X+9
AB²=3²+4²=25
①AP²=BP²：
X²+16=X²+6X+9
6X=7
X=7/6
P1（7/6,0）
②AP²=AB²：
X²+16=25
X²=9
X1 = - 3 (she), X2 = 3
P2（3,0）
③BP²=AB²：
（X+3）²=25
X+3=±5
X+3=5,X1=2
X+3=-5,X2=-8
P3（2,0）、P4（-8,0）
In fact, the P coordinates of the above points can be made by geometric method
(1) AB = BP = 5, so p is 5 units to the left of B, which is (- 8,0); or 5 units to the right of B, which is (2,0)
(2) If AB = AP, then Ao is the vertical bisector of BP, and O is the midpoint of BP, so p is (3,0)
(3) If AP = BP, then p is on the right side of the origin
In RT △ AOP, Ao = 4, ap-op = bp-op = ob = 3
If OP is x, then AP is x + 3
X²+4²=（X+3）²
6X=7
X=7/6
In addition, your question: √ (x? 2 + 4?) = | x + 3? Can remove the root sign and the absolute value sign as long as the square on both sides

2013 times of (Radix 2 + 1) and 2014 times of (Radix 2-1) were calculated

The original formula = (root 2 + 1) of 2013 times × (root 2-1) of 2013 times × (root 2-1)
=2013 times of [(Radix 2 + 1) × (Radix 2-1)] (Radix 2-1)
=2013 times of (2-1)
=1 × (radical 2-1)
=Radical 2-1

Let a = root sign (2012 / 2013), B = root sign (2013 / 2014) comparative size: A / b ﹣ 1

According to the question, a / B is greater than 0, so we can compare the square of (A / b) with the size of 1 (a ^ 2) / (b ^ 2) = (2012 * 2014) / 2013 ^ 2 = 0.999.72... < 1, so a / b < 1

Let a, B be real numbers, and a−5-2 5−a=b+4； (1) Find the value of a and B (2) Find the arithmetic square root of a-b

(1) According to the meaning of the title, a-5 ≥ 0 and 5-a ≥ 0,
A ≥ 5 and a ≤ 5 are obtained,
So, a = 5,
b+4=0，
The solution B = - 4;
（2）a-b=5-（-4）=5+4=9，
∵32=9，
The arithmetic square root of A-B 3

Given that the real number x, y satisfies (x-square-2013]) x (y-square-2013]) = 2013, what is 3 (x-square) - 2 (y-square) + 3x-3y-2012

√ (x ^ 2-2013) can be regarded as a solution of the equation y ^ 2-2xy + 2013 = 0, and the other solution of the equation is x + √ (x ^ 2-2013). Similarly, Y - √ (y ^ 2-2013) can be regarded as a solution of the equation x ^ 2-2xy + 2013 = 0. Two equations of the same type have the same solution, so the values of the two solutions are equal

Given that the real number x satisfies | 2013-x | + x-2014 = x under the root sign, find the square value of x-2013, be more careful, why is x greater than or equal to 2004?

Because the square root must be > = 0
Therefore, x-2014 > = 0, that is, x > = 2014
|2013-x|=x-2013
So the original formula: x-2013 + and x-2014 = X
It is concluded that x-2014 = 2013
Square of both sides: x-2014 = 2013
So x-2013 2 = 2014