X^2-2x+1/x^2-2x+2+x^2-2x+2/x^2-2x+3=7/6

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вопрос

To simplify the given equation, let's focus on each term separately:

(x^2 - 2x + 1) / (x^2 - 2x + 2): This simplifies to 1 since the numerator and denominator are the same.

(x^2 - 2x + 2) / (x^2 - 2x + 3): This fraction cannot be simplified further since the numerator and denominator are different quadratics.

Putting both terms together, we get:

1 + (x^2 - 2x + 2) / (x^2 - 2x + 3) = 7/6

Now, let's solve for x:

1 + (x^2 - 2x + 2) / (x^2 - 2x + 3) = 7/6
Multiply everything by 6 to get rid of the fractions:
6 + 6(x^2 - 2x + 2) / (x^2 - 2x + 3) = 7

Multiply through by the denominator (x^2 - 2x + 3) to clear the fractions:
6(x^2 - 2x + 3) + 6(x^2 - 2x + 2) = 7(x^2 - 2x + 3)
6x^2 - 12x + 18 + 6x^2 - 12x + 12 = 7x^2 - 14x + 21

Combine like terms:
12x^2 - 24x + 30 = 7x^2 - 14x + 21

Rearranging terms:
5x^2 - 10x + 9 = 0

Now, we have a quadratic equation that we can solve using the quadratic formula:
x = [-(-10) ± √((-10)^2 - 4 5 9)] / (2 * 5)
x = [10 ± √(100 - 180)] / 10
x = [10 ± √(-80)] / 10
x = [10 ± 2√20 i] / 10
x = (1 ± √5 i) / 5

Therefore, the solutions for x are:
x = (1 + √5 i) / 5 or x = (1 - √5 i) / 5