Let's solve the equation step by step:
x+12x−2−x−12x+2+2x(1−x)(x+1)=0
First, simplify each term individually:
1. Simplify
x+12x−2:
x+12(x−1)
2. Simplify
x−12x+2:
x−12(x+1)
Now the equation becomes:
x+12(x−1)−x−12(x+1)+2x(1−x)(x+1)=0
Next, find a common denominator for the fractions. The common denominator for
2(x−1),
2(x+1), and
(1−x)(x+1) is
2(x−1)(x+1).
Rewrite each fraction with the common denominator:
1.
x+12(x−1) becomes:
(x+1)(x+1)2(x−1)(x+1)=(x+1)22(x−1)(x+1)
2.
x−12(x+1) becomes:
(x−1)(x−1)2(x−1)(x+1)=(x−1)22(x−1)(x+1)
3.
2x(1−x)(x+1) becomes:
2x−(x−1)(x+1)=−2x(x−1)(x+1)=−2x(x−1)(x+1)
Now the equation is:
(x+1)22(x−1)(x+1)−(x−1)22(x−1)(x+1)−2x(x−1)(x+1)=0
Combine the fractions:
(x+1)2−(x−1)2−2x2(x−1)(x+1)=0
Simplify the numerator:
(x+1)2−(x−1)2−2x
Expand the squares:
(x2+2x+1)−(x2−2x+1)−2x
Combine like terms:
x2+2x+1−x2+2x−1−2x
4x
So the equation becomes:
4x2(x−1)(x+1)=0
Simplify the fraction:
2x(x−1)(x+1)=0
For the fraction to be zero, the numerator must be zero:
2x=0
x=0
Therefore, the solution to the equation is:
x=0
However, we should check if
x=0 is within the domain of the original equation. The original equation has denominators
2(x−1),
2(x+1), and
(1−x)(x+1). None of these denominators are zero when
x=0, so
x=0 is a valid solution.