Given the absolute value of x-3 + (y + 4) 2 + root sign, x + Y-Z = 0, find the value of (2x + y) Z power

Given the absolute value of x-3 + (y + 4) 2 + root sign, x + Y-Z = 0, find the value of (2x + y) Z power

|x-3|+(y+4)²+√(x+y-z)=0
They're all nonnegative numbers
∴x-3=0
X=3
y+4=0
y=-4
x+y-z=0
z=x+y=-1
(2x + y) Z power
=The - 1 power of 2
=1/2

If x, y are real numbers, and the root sign (x-2y-6) + (2x + y) 2 - 4 (2x + y) + 4 = 0, find the 2013 power of (Y in x)?

The root sign (x-2y-6) + (2x + y) 2 - 4 (2x + y) + 4 = radical (x-2y-6) + (2x + Y-2) 2 = 0, so there is x-2y-6 = 0, 2x + Y-2 = 0, so x-2y-6 = 0, 2x + Y-2 = 0. Therefore, x-2y-6 = 0, 2x + Y-2 = 0 can solve the two equations x = 2, y = 2

If y = under the root sign (1-2x) + the root sign (2x-1) + the root sign (x-1) 2, find the 2010 power of (4x-2y)

1-2x≥0,2x-1≥0
Therefore, 2x-1 = 0, x = 1 / 2
y=0+0+1/2=1/2
4x-2y=2-1=1
The answer is 1

We know: the absolute value of 2x + Y-3 + the Y power of (x-3y-5) ^ 2 = 0 x under the root sign=

Under the absolute value of 2x + Y-3 + root sign (x-3y-5) ^ 2 = 0
Then 2x + Y-3 = 0 x-3y-5 = 0
The solution is: x = 2, y = - 1
x^y=2^(-1)=1/2

Is Y1 = (root 2x-5) 2 and y2 = absolute value 2x-5 equal? Why? Is it that the nth power of root x is equal to y = x

Unequal
Square first, square root
That is √ A / = | a|
If the root sign is used first, then 2x-5 ≥ 0 is required
In this case, Y1 = 2x-5, there is no absolute value
The nth power of (the nth root of x) is X
N is an odd number, and the nth root of (x to the nth power) is X
N is an even number, and the nth root of (x to the nth power) = | x|

2 * (- 1) to the power of 2008 + the absolute value of - 8 divided by - 1 / 2 - Radix 64 / 81

2 * (- 1) to the power of 2008 + the absolute value of - 8 divided by - 1 / 2 - Radix 64 / 81
=2*(-1)^2008+³√-8÷│-1/2│-√64/81
= 2-2÷1/2-8/9
=2-4-8/9
=-26/9
Have a good time

Given that the real numbers x and y satisfy the square - 10x + root sign (y + 4) + 25 = 0 of X, then the value of (x + y) to the power of 2011? If x and y are real numbers and y = radical (x-1) + radical (1-x) + 2, find the value of (2 / 3x radical 9x + Radix 8xy) - (cube of radical x + radical 50xy)

x²-10x+25+√(y+4)=0
(x-5)²+√(y+4)=0
x=5,y=-4
(x+y)^2011=(5-4)^2011=1

If the real number XY satisfies x + 1 + (Y-5) 2 under the radical sign, then the power value of XY is?

X + 1 + (Y-5) 2 = 0 under radical
x+1=0,y-5=0
x=-1,y=5
therefore
The Y power of X
=The fifth power of (- 1)
=-1

Let XY be a real number and satisfy the condition that the root sign 1 + X - (Y-1). The root sign 1-y = 0, try to find the 2009 power of X - the 2009 power of Y

√(1+x)+(1-y)√(1-y)=0
Greater than or equal to 0 under root sign
1-y>=0
And the arithmetic square root itself is greater than or equal to 0
So √ (1 + x) > = 0
√(1-y)>=0
So √ (1 + x) > = 0
(1-y)√(1-y)>=0
If the sum is 0, then all of them are equal to 0
So 1 + x = 0, 1-y = 0
x=-1,y=1
Original formula = (- 1) ^ 2009-1 ^ 2009
=-1-1
=-2

If x, y are real numbers and | x + 2|+ If Y-3 = 0, then the value of (x + y) 2010 is___ .

According to the meaning of the title, x + 2 = 0, Y-3 = 0,
X = - 2, y = 3;
So (x + y) 2010 = 1
So the answer is: 1