Given that the real numbers x and y satisfy the root sign (x-1) + y ^ 2 = 4y-4, find the value of (X-Y) ^ 2012

Given that the real numbers x and y satisfy the root sign (x-1) + y ^ 2 = 4y-4, find the value of (X-Y) ^ 2012

The root sign (x-1) + y ^ 2 = 4y-4 √ (x-1) + (y-4y + 4) = 0 √ (x-1) + (Y-2) 2 = 0, because √ (x-1) and (Y-2) 2 are ≥ 0 respectively. In order to √ (x-1) + (Y-2) 2 = 0, it must be √ (x-1) and (Y-2) 2 = 0, i.e., X-1 = 0, x = 1y-2 = 0, y = 2 (X-Y) ^ 2012 = (1

If the real number x, y satisfies y = radical 2x-1 + radical 1-2x + 1 / 3, find the value of the algebraic expression x ^ 2-2xy + y ^ 2 Can you give me a more detailed answer

2x-1 and 1-2x are opposite numbers to each other, because the number under the root must be non negative, so x = 1 / 2, y = 1 / 3;
The value of the algebraic expression is 1 / 36

If x, y are real numbers and Y is less than the radical X-1 + the radical 1-x + 1 / 2, find the absolute value 1-y of Y-1

The root sign X-1 + root sign 1-x gives x = 1, so y is less than 1 / 2;
1-y is greater than 1 / 2,
The absolute value of Y-1 is 1-y = 1

If x.y is a real number and Y is less than the root sign (x-1) + the root sign (1-x) plus 1 / 2, the absolute value (1-y) is divided by (Y-1) if x.y

The title is like this: No: y

The radical of (y) + the radical of (y) + 1

1-8x>=0
8x-1>=0
∴x=1/8
When x = 1 / 8, y = 1 / 2
∴x/y=1/4
y/x=4
Radical (x in y + y + 2 in x) - (x in y + Y-2 in x)
=√(1/4+4+2)-√(1/4+4-2)
=5/2-3/2
=1

Given that y = radical 1-8x + Radix 8x-1 + 1 / 2, find the value of the algebraic formula X of radical y + y of X + 2-of radical y + X + Y-2 of X

1-8x≥0
x≤1/8
8x-1≥0
x≥1/8
So x = 1 / 8, y = 0 + 0 + 1 / 2 = 1 / 2
x/y=1/4 y/x=4
simple form
=√（1/4+4+2）-√(1/4+4-2)
=√25/4-√9/4
=5/2-3/2
=1

The second root of grade 9 is known: y = radical 1-8x + Radix 8x-1 + 1 / 2, and find the algebraic root sign x / y + Y / x + 2

Because y is meaningful, it must be 1-8x > = 0, 8x-1 > = 0, 8x-1 > = 0 = 0, 8x-1 > = 0 = = > x = 1 / 8 = = > x = 1 / 8 = > x = 1 / 8, y = 1 / 2, so √ (x / y + Y / x + 2) = (1 / 8) / (1 / 2) + (1 / 2) / (1 / 8) + (1 / 8) + (1 / 2) / (1 / 8) + 2) = (25 / 4) because x > 0, Y > 0 is inevitable √ (x / y + y + X + 2) > 0, so, so √ (x / y + y + Y / x + 2 + 2) = 5, therefore, therefore, therefore, therefore, it's √ (x / y + y + / 2

1. Given that: y = radical (1-8x) + radical (8x-1) + 1 / 2, find the value of the algebraic radical (x / y + Y / x + 2) - radical (x / y + Y / X-2)

Since both radical (1-8x) and radical (8x-1) exist, there are 1-8x > = 0 and 1-8x

Y is known= 1−8x+ 8x−1+1 2, then algebraic expression X Y+y The value of X + 2 is () A. ±5 Two B. -5 Two C. 5 Two D. Can't be sure

∵y=
1−8x+
8x−1+1
2 1-8x ≥ 0, 8x-1 ≥ 0,
∴x=1
8，y=1
Two
Algebraic expression
X
y+  y
x+2 
=
One
4+4+2 
=
Twenty-five
Four
=5
2．
Therefore, C

The position of real numbers a, B, C on the number axis is shown in the figure | a | = | B | simplification, | a | + | a + B | - root sign (C-A) square - 2 * root sign C square Number axis_____________ c a 0 b

|a|=|b|
From the number axis: a