If the absolute value of the radical X-1 + Y-3 = 0, what is the equivalent of X * XY / 2Y

If the absolute value of the radical X-1 + Y-3 = 0, what is the equivalent of X * XY / 2Y

The root number is greater than zero, so x = 1, y = 3
Replace it with (2 root sign 1) * (root 3) / (root 6) = root 2

Given that y is less than the root (x-1) + heel (1-x) + 1 / 2, simplify the absolute value (2y-1) - radical (y2-2y + 1)

The answer of simplification is: - y according to the question, we should first find the value range of X that satisfies the condition, that is: (x-1) ≥ 0 and (1-x) ≥ 0, we can find x ≥ 1 and X ≤ 1. If the two conditions are satisfied at the same time, x can only be equal to 1, so substituting x = 1 into the original condition can obtain y ＜ half

The curve represented by the equation y = 2 + root sign (2-x 2) is A parabola B circle C-ray D semicircle

B
y=2+√（2-x²） y≥2
（y-2）²=2-x²
(y-2)²+ x²=2
Therefore, it is a semicircle with the center (0,2) and radius √ 2

Equation X-1= The curve represented by 1 − (Y − 1) 2 is () A. A circle B. Two semicircles C. Two circles D. Semicircle

∵ equation X-1=
1 − (Y − 1) 2 is equivalent to (x-1) 2 + (Y-1) 2 = 1 (x ≥ 1),
The curve represented is a half circle
Therefore, D

Draw the curve of the following equation: (1) y = 9-x ^ 2 under root sign (2) y = 4x-x ^ 2 under root sign

(1) By moving the square of both sides, we can get x ^ 2 + y ^ 2 = 9 (Y > = 0)
So the curve is the top half of a circle with radius 3, centered at the origin, including two points on the x-axis
(2) The square of both sides is 4x-x ^ 2 = (3-y) ^ 2, and then (X-2) ^ 2 + (Y-3) ^ 2 = 4 (y)

Given that the 2-radical sign 5 is a heel of the quadratic equation of one variable x? - 4x + C = 0, then the other following of the equation is

2 + radical 5
Because according to Veda's theorem: X1 + x2 = 4
X2 = 4-x1 = 4 - (2-radical 5) = 2 + radical 5

Given the absolute value of 2-x under the root sign + the absolute value of 1-y under the root sign = 0 and the absolute value of X-Y = Y-X, find the value of X + y i 'm sorry

√(2-|x|)+√(1-|y|)=0
Then 2 - | x | = 0 and 1 - | y | = 0
∴|x|=2,|y|=1
x=±2,y=±1
∵|x-y|=y-x
∴y-x≤0,y≤x
∴{x=2,y=2
or
｛x=2,y=-2
∴x+y=4
Or x + y = 0

It is known that there is a point P (- 1,2) in the square + y-square = 8 of the circle (x + 1) and the straight line AB passes through the point P: when the absolute value of the chord length AB is equal to two times the root sign seven, find the inclination of ab

The chord length is 2 √ 7, the radius of the circle is 2 √ 2, the center distance is 1, and the center of the circle is (- 1,0)
1. If the slope of the straight line does not exist, then the line is x = - 1, which is not satisfied;
2. If the slope of the straight line exists, let the straight line be y = K (x + 1) + 2, then the distance from the center of the circle to the straight line = | 2 | / √ [1 + k?] = 1, the solution is k = ±√ 3, and the inclination angle is 60 ° or 120 °

If the square root of X + Y-1 + (Y-2) + the absolute value of Z-1 = 0, find the value of X + y + Z

If x + Y-1 + the square of (Y-2) + the absolute value of Z-1 = 0
Namely
x+y-1=0,（y-2）=0,z-1=0
y=2,z=1
x=1-y=1-2=-1
therefore
x+y+z=-1+2+1=2

The straight line L: y = x + B and the curve C: y= If 1-x2 has two common points, then the value range of B is () A. - 2＜b＜ Two B. 1≤b＜ Two C. -1≤b≤ Two D. 1≤b≤ Two

Draw the corresponding figures according to the meaning of the title, as shown in the figure
When the line L is tangent to the circle, the distance from the center of the circle (0, 0) to y = x + B is d = r = 1,
I.e|
2 = 1, the solution is: B=
2 or B=-
2 (give up)
When the line L passes through (- 1,0), substitute (- 1,0) into y = x + B,
The result is: B = 1,
When the line L and the curve C have two common points, the range of B is 1 ≤ B ＜
2，
Therefore, B