Solving (3) x1-x2 with known equation x ^ 2-3x + 1 = 0

Solving (3) x1-x2 with known equation x ^ 2-3x + 1 = 0

x1+x2=3
x1x2=1
(x1-x2)²
=(x1+x2)²-4x1x2
=9-4
=5
So x1-x2 = ± √ 5

Let's first read the following passage. Here is a class of equations and their solutions: the solutions of X-1 / x = 1 and 1 / 2 are X1 = 2, X2 = - 1 / 2, X-1
The solution of X-1 / x = 2 and 2 / 3 is X1 = 3, X2 = - 1 / 3
Observe the above equation and its solution, guess the solution of equation X-1 / x = 10 and 10 / 11, and write the test!

For the most common form, X-1 / x = n + n / (n + 1)
By solving this equation, we can get that X1 = n + 1, X2 = - 1 / (n + 1)
Taking n = 10, we get: X1 = 11, X2 = - 1 / 12

Read the following examples: solve the equation x ^ 2 - | x | - 2 = 0, (1) when x ≥ 0, the original equation is changed to x ^ 2-x-2 = 0, and the solution is X1 = 2, X2 = - 1 (unqualified)
Read the following examples:
Solve the equation x ^ 2 - | x | - 2 = 0,
(1) When x ≥ 0, the original equation is reduced to x ^ 2-x-2 = 0,
The solution is X1 = 2, X2 = - 1 (not qualified to be rejected)
When x < 0, the original equation is reduced to x ^ 2 + X-2 = 0
The solution is X1 = 1 (rounding off), X2 = - 2
The solution of the original equation is X1 = 2, X2 = - 2
Please refer to the appeal method to solve the equation: x ^ 2 - | X-1 | - 1 = 0
（2）∵x^2=|x|^2
The original equation can be changed to: | x | ^ 2 - | x | - 2 = 0,
Take | x | as a whole and get | x | 1 = 2, | x | 2 = - 1 (rounding off)
∴x1=2,x2=-2
Please refer to the appeal method to solve the equation: x ^ 2-2x - | X-1 | - 5 = 0
Help me. If you can do it, try to answer my questions

Here are some tips:
(1) When x ≥ 1, the original equation is changed to x ^ 2-x = 0
When x