Given (a + 2) + (B-2 / 1), find the product of a's 2006 power and B's 2007 power

Given (a + 2) + (B-2 / 1), find the product of a's 2006 power and B's 2007 power

|A+2|+|B-1/2|=0
A=-2,B=1/2
The product of power 2006 of a and power 2007 of B
=(-2)^2006*(1/2)^2007
=(-2*1/2)^2006*1/2
=1/2

The power of 2008 of (2-radical 3) * (2 + radical 3) = ()

The power of 2008 of (2-radical 3) and 2009 power of (2 + radical 3)
=2008 power of [(2-radical 3) (2 + radical 3)] (2 + radical 3)
=2008 power of (4-3) * (2 + radical 3)
=2 + radical 3

The radical of - 2 is 3 minus the power of 2

1 / 12 (1 / 12)

If x and y are real numbers and y = radical x-3 + radical 3-x + 8, find the square root of X + 3Y Here, here, here, standing, standing, standing

Greater than or equal to 0 under root sign
So x-3 > = 0,3-x > = 0
The opposite is the sum of x-3
At the same time, if it is greater than or equal to 0, it is equal to 0
x-3=0,x=3
So root x-3 = 0, radical 3-x = 0
y=0+0+8=8
x+3y=27
So the square root of X + 3Y = ± 3 √ 3

Given that the real number x, y satisfies y = radical X-1 + Radix 1-x + 8, find the square root of radical x + y

∵1－x≥0,x≤1
x－1≥0,x≥1
∴ x＝1
y＝8
∴ √﹙x＋y﹚＝√﹙1＋8﹚＝3
The square root of √ (x + y) is ±√ 3

It is known that X and y are real numbers, and Y= x−2+ 2 − x + 4, then the square root of YX is______ .

∵ negative numbers cannot be squared,
Qi
x−2≥0
2−x≥0 ，
∴x=2，y=4，
∴yx=42=16，
∴±
16=±4，
So the answer is: ± 4

It is known that Y-1 and 3-2x under the cubic root sign are opposite to each other, and the square root of X-Y + 4 is itself. Find the value of X and y

(Y-1) ^ (1 / 3) = - (3-2x) ^ (1 / 3), that is, Y-1 = - (3-2x), that is - 2x + y = - 2
The square root of X-Y + 4 is itself, that is, X-Y + 4 = 0 or 1
Two equations are obtained
-2X + y = - 2, X-Y + 4 = 0, x = 6, y = 10
-2X + y = - 2, X-Y + 4 = 1, x = 5, y = 8
Therefore, there are two groups of solutions

Given that (x + Y-1) square and root sign 2x-y + 4 are opposite numbers to each other, find the square root of X + y square

Square (x + Y-1) and root sign 2x-y + 4 are opposite numbers
ν (x + Y-1) square + root sign 2x-y + 4 = 0
∴x+y-1=0
2x-y+4=0
x=-1
Y=2
∴x²+y²=5
The square root of X + y = ± √ 5

Known X − y + 3 and If x + y − 1 is opposite to each other, find the square root of (X-Y) 2

Solution
X − y + 3 and
X + y − 1 is opposite to each other,
Qi
x−y+3+
x+y−1=0，
 X-Y + 3 = 0 and X + Y-1 = 0,
X = - 1, y = 2,
∴（x-y）2=（-1-2）2=9，
∵（±3）2=9，
The square root of (X-Y) 2 is equal to ± 3

On the problem of the square root of mathematics! It is known that x + Y-3 and x-y-1 under the root sign are opposite to each other. How much is X-Y?

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 x + Y-3 = 0
x-y-1=0
So X-Y = 1