If the absolute value of X-2 and the radical 2Y + 1 are opposite to each other, then XY=

If the absolute value of X-2 and the radical 2Y + 1 are opposite to each other, then XY=

x-2=0
2y+1=0
x=2,y=-1/2
xy=-1

If x, y satisfy the root sign Π - x + radical sign x - Π + absolute value 1-2y = 5, then what is x = and what is y?

Greater than or equal to 0 under root sign
So Π - x > = 0, x = 0, x > = Π
At the same time
So x = Π
Then Π - x = 0, X - Π = 0
So 0 + 0 + | 1-2y | = 5
1-2y=±5
2Y = 1 ± 5 = - 4 or 6
So x = Π, y = - 2 or 3

If the radical of (y + 2x) - 2x = 2x-2x, then (y + 2x) = (2x-1) = (x-1) = (x-1) = (x-1) = (x-1) = (y + X-2) = (x-1) = (x-1) = (X-2) = (X-2) =! If the absolute value of (x-2y + 1) + root sign (2x-y-5) = 0, then x + y = how much

According to the meaning of the title, x-2y + 1 = 0
2x-y-5=0
By subtracting the two formulas, x + y = 6
It can also solve the equations and find the value of X, y, but it is troublesome

If the radical (x + 2y-5) + absolute value (2x-3) = 0, then x =, y=

Radical (x + 2y-5) + absolute value (2x-3) = 0
So 2x-3 = 0
x+2y-5=0
X = 3 / 2, y = 7 / 4

Given the absolute value of m-2 of the first order function y = (m-2) x to the power of - m, the value of M is calculated through the second, third and fourth quadrants

Because it's a function of order
So | m-2 | = 1
So M1 = 3, M2 = 1
Because it passes through the second, third and fourth quadrants
So: m-2

Given the absolute value of M of the first order function y = 2x-2 power + m-2 image, through the 1,2,3 quadrants, to find the expression of this linear function

According to the meaning of the title
The absolute value of M of function y = 2x to the power of - 2 + m-2 is a function of order 1
So the absolute value of m to the power of - 2 = 1
So m = ± 1
Therefore, the analytic formula of the first order function is:
y＝2x－1
or
y＝2x－3
But these two function images do not pass through the first, second and third quadrants
So there is no solution
(please check if there is any transcription error in the title)

When n is taken, the absolute value of n of y = (n-1) x to the power of - 2 is an inverse proportional function. What quadrant is his image in? How does y change with the increase of X in each quadrant

The - 2 power of | n | = - 1
Square of n = 1
N = 1 or - 1
∵n-1≠0
∴n≠1
∴n=-1
n-1=-1-1=-2
The image is in the second and fourth quadrants
In each quadrant, y increases with the increase of X

Given that the point P (x.y) is on the image of the function y = 1 / x? 2 + radix-x, which quadrant should the point p be in the rectangular coordinate system? The steps are clear

Because there is a root - x, so - x must be ≥ 0. So x ≤ 0. And because 1 / x 2 > 0, root - x > 0, so sum > 0, that is, Y > 0, so it is in the second quadrant

If the image with positive scaling function y = (A-2) x passes through the first and third quadrants, what is the result of (A-1) 2 under the root sign

The image with y = (A-2) x passes through the first and third quadrants
a-2>0
A>2
√(a-1)²=|a-1|
a-2>0
a-1=(a-2)+1>1
√(a-1)²=(a-1)

What is the distance between the function image of the quadratic function y = - x ^ 2 + 2 radical sign 3x + 1 and the two intersection points of X axis?

Then y = 0 on the x-axis
So - x 2 + 2 √ 3x + 1 = 0
x1+x2=2√3
x1x2=-1
So (x1-x2) 2
=(x1+x2)²-4x1x2
=16
So the distance = | x1-x2| = 4