X2 + Y2 + Z2 = 1 to find the value range of X + y

X2 + Y2 + Z2 = 1 to find the value range of X + y

x^2+y^2+z^2=1
=>x^2+y^2=1-z^2
=>(x+y)^2-2xy=1-z^2
(x+y)^2=x^2+y^2+2xy>=4xy
=>xy=(x+y)^2-(x+y)^2/4*2=(x+y)^2/2
=>(x+y)^2

Given a = 2x3 XYZ, B = Y3-Z2 + XYZ, C = - x2 + 2Y2 XYZ, and (x + 1) 2 + / Y-1 / + / Z / = 0, find the value of a - (2b-3c)

Because (x + 1) 2 + / Y-1 / + / Z / = 0, and (x + 1) 2 ≥ 0 / Y-1 / ≥ 0 / Z / ≥ 0
So x = - 1, y = 1, z = 0
So a = - 2, B = 1, C = 1
So a - (2b-3c) = - 2 - (2-3) = - 1

Must the generatrix of a cylinder be perpendicular to the plane formed by the guide line?
Is the generatrix of a cylinder unique? If the generatrix is z-axis, can a line parallel to z-axis be the generatrix of a cylinder? Can a line not parallel to z-axis be the generatrix of a cylinder?

1. No
2. Not unique. The cylinder is made up of a cluster of generatrix, but the generatrix must be in the plane. That is to say, the generatrix must at least intersect the guide line~
Can a line not parallel to the Z axis be a generatrix of a cylinder

It is proved that the surface f (x / L-Y / m.y / M-Z / n.z/n-x / L) = 0 is a cylinder whose generatrix is parallel to the straight line X / L = Y / M = Z / n

Suppose that the point (x, y, z) is on the surface F
So it's easy to verify that (x + LT, y + MT, Z + NT) is also on the surface
So (L, m, n) is a translation direction of the surface, so it is a cylinder

If a plane passes through a straight line 3x + 4y-2z + 5 = 0, x-2y + Z + 7 = 0, and has an intercept-3 on the z-axis, the equation for it is obtained

Let the plane equation: (3x + 4y-2z + 5) + K (x-2y + Z + 7) = 0, substitute x = 0, y = 0, z = - 3 to get k = - 11 / 4, and the plane equation is: x + 38y-19z-57 = 0

On the overturning of the volume of revolution in Higher Mathematics
Excuse me, teachers, how to deduce the formula of y-axis rotation?

If there are only two points on the circle (x-3) 2 + (y + 5) 2 = R2 and the distance from the two points to the straight line 4x-3y = 2 is equal to 1, then the value range of radius R is ()
A. （4，6）B. [4，6）C. （4，6]D. [4，6]

∵ the distance from the center of the circle P (3, - 5) to the straight line 4x-3y = 2 is equal to | 12 − 3 · (- 5) − 2 | 16 + 9 = 5, from | 5-r | 1 to get & nbsp; & nbsp; 4 | R | 6, so select a

(x + y-xy) / (x + y + 2XY) = (y + z-2yz) / (y + Z + 3yZ) = (Z + x-3zx) / (Z + X + 4zx), and 2 / x = 3 / y = 1 / Z, then XYZ =?

Let 2 / x = 3 / y = 1 / Z be K
x=2k,y=3k,z=k
Substituting 1-k / 1 + 2K = 4-6k / 4 + 9K = 3-6k / 3 + 8K
Calculate K to get k = - 1
x=-2,y=-3,z=-1
xyz=-6

Given (x + y-xy) / (x + y + 2XY) = (y + z-2yz) / (y + Z + 3yZ) = (Z + x-3zx) / (Z + X + 4zx) and 2 / x = 3 / Y-1 / Z, then XYZ =?

Let (x + y-xy) / (x + y + 2XY) = (y + z-2yz) / (y + Z + 3yZ) = (Z + x-3zx) / (Z + X + 4zx) = k, then
The 1 / x + 1 / x + 1 / y is as a whole to solve the 1 / x + 1 / y as a whole to get 1 / x + 1 / y = (1 + 2K) / (1-k) / (1-k) / (1-k) (x + y-y-xy) / (x + y + y-xy) / (x + y + X + 2XY) = (1 / x + 1 / Y-1 / Y-1 / (1 / x + 1 / Y-1 / (1 / x + 1 / x + 1 / 1 / 1 / x + 1 / y + 1 / y + 1 / Y + 1 / Y-1 / (1 / x + 1 + 3K / 2) / (1-k) / (1-k) 1 / x = (1 / x = (1 + 3K / 3K / 2) / (1-k) / (1-k) / (1-k) (1-k) (1-k) (1-k) (1 / 1-k) 1 / y = (1 / k) (1 x = y = z = - 4, XYZ = - 64

Given that | x | = 1, | y | = 2, | Z | = 3, and xy0, try to find the value of (x + y + Z) × (XY + YZ)

Because xy0
So z = - 3
When x = 1, y = - 2
For - 16
When x = - 1, y = 2
16