Given x + 2 / X / = - X-2 / x, then the value range of X is?

Given x + 2 / X / = - X-2 / x, then the value range of X is?

It is simplified that x (x + 2) = - / X / (x + 2), so the square of both sides of the value range of X is obtained, and X is negative infinity to 0

Let XYZ = 1, find the value of XXY + X + 1 + YYZ + y + 1 + Zzx + Z + 1

The original formula = XXY + X + 1 + xyxyyz + XY + X + Zzx + Z + 1, = XXY + X + 1 + xy1 + XY + X + zxyzx · XY + ZXY + XY, = XXY + X + 1 + xyxy + X + 1 + 1xy + X + 1, = XY + X + 1xy + X + 1, = 1

1 / A-1 / 3

1/a-1/3=3/3a-a/3a=(3-a）/3a

On the transformation of the main term of some middle two formulas and the simple algebraic fraction
Simplify the following algebraic fractions
1.u-1/u+1+u+1/u-1
2.u-1/u+1-u+1/u-1
1-x / a = x / b
S = n / 2 (a + L)
S = ut + 1 / 2at square
X (Y-2) = y + 1
F = (9 / 5) C + 32
A = 1 / 2 × (a + b) × H
These are not. Please ask. Write out the answer process. Thank you

On the calculation of identical deformation by fraction
It is known that x, y and Z are three unequal real numbers, and X + (1 / y) = y + (1 / z) = Z + (1 / x)
Verification: X & sup2; Y & sup2; Z & sup2; = 1
Thank you very much!

It is known that X-Y = 1 / y + 1 / z = (Y-Z) / YZ
So YZ = (Y-Z) / (X-Y)
The rest are also pushed in this way
Finally, we get the original formula = XY * YZ * ZX = 1

Because the solution fraction and () are mutually opposite deformations

Multiplicative distributive law

The second volume of the first grade of junior high school (PEP) is a problem of fractional transformation,
It is known that x = (Y-1) / (y + 1) is expressed as y by containing algebraic expression, then y =?
That's using algebra with X to represent y. I'm missing an X

Multiply both sides by Y + 1 to get x (y + 1) = Y-1; XY + x = Y-1; (x-1) y = - 1-x; y = (x-1) / (- x-1)

The fraction-a / A-B can be transformed into

A / B-A or - A / a-b

Mixed operation of one question fraction
[x-1/x²-x-2]÷x²-2x+1/2-x÷[1/x²+x]

[x-1/x²-x-2]÷x²-2x+1/2-x÷[1/x²+x]
=[x-1/(x-2)(x+1)]÷(x-1)^2/2-x*(x+1)x
=(x-1)(2-x)*(x+1)x/(x-2)(x+1)(x-1)^2
=x/1-x
=1/(x-1)-1

Help to solve math problems (second day fraction mixed operation)
x/x-1-2/x=1
(one-third minus one-fifth) times one-fifth of the square outside the brackets divided by the absolute value of negative one-third plus the 0 power of (root sign minus two) plus negative one-quarter of the square outside the brackets multiplied by the 2007 power of 4

X / (x-1) - 2 / x = 1 is this right? Multiply both sides by X (x-1) x ^ 2-2 (x-1) = x (x-1) x ^ 2-2x + 2 = x ^ 2-xx = 2 to test, and establish [(1 / 3-1 / 5) * (1 / 5)] ^ 2 / | - 1 / 3 | + (√ 3-2) ^ 0 + (- 1 / 4) ^ 2006 * 4 ^ 2007 = [(2 / 15) * (1 / 5)] ^ 2 / (1 / 3) + 1 + (1 / 4) ^ 2006 * 4 = (2 / 75) ^ 2 / (1 / 3)