What is the curve represented by the equation | X-1 | = root sign (1 - (Y-1) ^ 2)?

What is the curve represented by the equation | X-1 | = root sign (1 - (Y-1) ^ 2)?

|X-1 | = radical (1 - (Y-1) ^ 2)
(x-1)^2=1-(y-1)^2 1-(y-1)^2≥0
(x-1)^2+(y-1)^2=1 (y-1)^2≤1
(x-1)^2+(y-1)^2=1 0≤y≤2
Represents a circle with (1,1) as its center and 1 as its radius

What is the curve represented by the equation | x | - 1 = root sign (1 - (Y-1) ^ 2)? Answer: from lxl-1 = root [1 - (Y-1) square], lxl-1 > 0, so when x > 1 or x1, (x-1) square + (Y-1) square = 1, the equation represents the right semicircle

Answer: from lxl-1 = root [1 - (Y-1) square], lxl-1 > 0, so when x > 1 or x1, (x-1) square + (Y-1) square = 1, the equation represents the right semicircle

Find the curve of equation (x + Y-1) root sign (x-y-2) = 0

The original equation is equivalent to x-y-2 = 0, or x + Y-1 = 0 under the condition of x-y-2 > 0, and the curve is composed of x-y-2 = 0 and X + Y-1 = 0 under the line x-y-2 = 0

What curve does equation (x + Y-1) root sign (x-1) = 0 represent

Firstly, the value range of the independent variable is determined. According to the fact that (x-1) is in the root sign, X is greater than or equal to 1;
Then, in order to make this equation hold, we must x + Y-1 = 0 or radix X-1 = 0, that is, y = 1-x or x = 1;
Combined with the value range of X, the curve contains two parts: straight line x = 1 and ray y = 1-x (x is greater than or equal to 1)

Y = the square of root 3x-x divided by the absolute value - 1 of X-1 to find the function definition domain

The solution is that 3x-x ^ 2 ≥ 0 and / X-1 / - 1 ≠ 0
That is, x ^ 2-3x ≤ 0 and / X-1 / ≠ 1
X ≤ 0 and X ≤ 0
That is, 0 < x < 2 or 2 < x ≤ 3
So the definition domain of function is {X / 0 ＜ x ＜ 2 or 2 ＜ x ≤ 3}

The definition domain of the 0 th power of (x-1) of the absolute value of X under the root sign is kuai kuai kuai

0 to the power of 0 is meaningless
So X-1 ≠ 0
x≠1
The root is greater than or equal to 0 and the denominator is not equal to 0
So | x | - x > 0
|x|>x
So X

Find the value range of the independent variable x in the following functions: y = 3x root 2, y = x divide root 1-x. y = root x + 1 / 3. Y = absolute value of X divided by X-1

Any value of X in question 1
The rest of the questions are one way
The root must be > = 0
The denominator cannot be equal to
So the answer comes out
X-3
X

Find the definition field of the following function y = root (X-2) + (X-5) x y = (- 3x-10) zero power + root sign (6-absolute value x)

y=√(x-2)+x/(x-5)
x-2>=0,x>=2
And X-5 ≠ 0, X ≠ 5
Definition domain [2,5) ∪ (5, + ∞)
y=(-3x-10)^0+√(6-|x|)
-3x-10≠0,x≠-10/3
6-|x|>=0
|x|

The following relations are as follows: y = 3x + 2; x2-y2 = 1, y = radical X; y = the absolute value of X; the absolute value of Y is equal to X; where y is a function of X, there are ()

The following relations are as follows: y = 3x + 2; x2-y2 = 1, y = the absolute value of X; y = the absolute value of X; the absolute value of Y is equal to X; where y is the function of X (y = 3x + 2; y = the absolute value of X)

If the absolute values of the radical x minus 1 and 2Y plus 4 are opposite to each other, then x plus y is equal to

The absolute values of root x minus 1 and 2Y + 4 are opposite to each other
Then the root sign (x-1) + |2y + 4| = 0
Then X-1 = 0,2y + 4 = 0
Then x = 1, y = - 2
X+Y=-1