It is known that the fractional part of 5 + radical 7 is a, the integer part is m, the decimal part of 5-radical 7 is B, and the integer part is n. find the value of (a + b) ^ 2012 Mn

It is known that the fractional part of 5 + radical 7 is a, the integer part is m, the decimal part of 5-radical 7 is B, and the integer part is n. find the value of (a + b) ^ 2012 Mn

∵ 5 + the decimal part of Radix 7 is a, the integer part is m, the decimal part of 5-radical 7 is B, and the integer part is n
∴a=5+√7-7,m=7,b=5-√7-2,n=2
∴（a+b）^2012-mn=﹙5+√7-7+5-√7-2﹚^2012-7×2=1-14=-13

It is known that XY is opposite to each other, AB is reciprocal to each other, and the absolute value of C is equal to 2, - 3 is a square root of Z. find the square of X - the square of Y + the square of C / (root AB)- Radical Z

If XY is opposite to each other, then x + y = 0
If AB is reciprocal to each other, then AB = 1
If the absolute value of C is equal to 2, then C = ± 2
If - 3 is a square root of Z, then there is √ z = - 3, so z = 9
x^2 -y^2 + c^2/√ab
=(x+y)(x-y) +c^2/√ab
=0 + (±2）^2 /1 = 4

Given that the absolute value of the cubic root sign x minus 4 plus the square root of Z minus 3 plus y minus 2Z, the square root of 1 is equal to 0. Find the cube root of X + (y * y * y) + (Z * Z * z)

Because the root and absolute value are both > = 0 3x-5y = 0, x-3y + 3 = 0, y = 9 / 4, x = 15 / 4 2x-2 / 3Y = 30 / 4-3 / 2 = 6, square root is root 6, and - root sign 6 is very good. You can try it
At the end of September 11, 2011, 12:09:22

Given that XY is reciprocal to each other, C and D are opposite to each other, the absolute value of a is 3 and the arithmetic square root of Z is 5. Find the value of 4 times the root of (c + D) + XY + a fraction

The absolute value of a is 3 and the arithmetic square root of Z is 5
So xy = 1
c+d=0
a=±3
√z=5
4 times the value of the root Z of (c + D) + XY + A
=4×0+1+5/(±3)
=1±5/3
=-2 / 3 or 8 / 3

It is known that the real number x, y, Z satisfies (absolute value 4x-4y + 1) + (1 / 3 radical 2Y + Z) + (Z squared - Z + 1 / 4) is equal to 0. Find the square of (y + Z) × X Not good, general. Just have a look. Thank you. There is a reward

(absolute value 4x-4y + 1) + (1 / 3 radical 2Y + Z) + (square of Z - Z + 1 / 4) = 0,
Is it an absolute value (4x-4y + 1) + 1 / 3 radical (2Y + Z) + (Z squared - Z + 1 / 4) = 0?
Absolute value (4x-4y + 1) > = 0,
1 / 3 (radical 2Y + Z) > = 0, (Radix 2Y + Z) > = 0,
(Z squared - Z + 1 / 4) = (Z-1 / 2) 2 > = 0,
Absolute value (4x-4y + 1) + 1,3 radical sign (2Y + Z) + (square of Z - Z + 1 / 4) = 0,
So:
Absolute value (4x-4y + 1) = 0,... (1)
(radical 2Y + Z) = 0, (2Y + Z) = 0... (2)
(Z-1/2)²=0,...(3)
From (3), z = 1 / 2
Y = - 1 / 4, substituting (1): 1
Absolute value (4x-4y + 1) = absolute value (4x + 1 + 1) = absolute value (4x + 2) = 0,4x = - 2, x = - 1 / 2,
（y+z）x²=(-1/4+1/2)(-1/2)²=1/16

It is known that the real numbers x, y, Z satisfy (absolute value 4x-4y + 1) + (1,3 radical 2Y + Z) + (Z squared - Z + 1,4) is equal to 0 It's not easy to fight. Just make do with it. You will be rewarded! Find the square of (y + Z) × X

|4x-4y+1|+1/3*√(2y+z)+(z^2-z+1/4)=0
|4x-4y+1|+1/3*√(2y+z)+(z-1/2)^2=0
be
4x-4y+1=0
2y+z=0
z-1/2=0
The solution
z=1/2
y=-1/4
x=-1/2

Known X + 2 + | Y-3 | = 0 The value of x2 + Y2 + 3

A kind of
X + 2 ≥ 0, | Y-3 | ≥ 0 and
x+2+|y-3|=0，
Qi
x+2=0，|y-3|=0，
∴x+2=0，y-3=0，
X = - 2, y = 3,
When x = - 2, y = 3,
x2+y2+3=
(-2)2+32+3=
16=4．
So the answer is: 4

If the absolute value of x-radical 3 + the square of (y + 3 / 3) equals 0, then the square of (XY) 2005 is equal to?

If one is greater than 0, the other is less than 0
So both are equal to zero
So x - √ 3 = 0, y + √ 3 / 3 = 0
x=√3,y=-√3/3
So xy = - 1
So (XY) 2005 power = - 1

It is known that x, y are opposite numbers to each other, a and B are reciprocal to each other, the absolute value of C is equal to 5, and - 3 is a square root of Z. we can find the value of x 2 + y 2 - √ Z / C 2

It is known that x, y are opposite numbers to each other, a and B are reciprocal to each other, the absolute value of C is equal to 5, and - 3 is a square root of Z,
x+y=0
ab=1
c=±5
Z=9
x²-y²-√z／c²
=0-√9/25
=-3/25

It is known that XY is opposite to each other, AB is reciprocal to each other, the absolute value of C is equal to 5, - 3 is a square root of Z, and the value of ab-x + C 2 + Z-Y is calculated

From the meaning of the title: ab = 1, x + y = 0, C ^ 2 = 25, z = (- 3) ^ 2 = 9
Therefore:
ab-x+c²+z-y
=1-（x+y）+c²+z
=1-0+25+9
=35