(x^2−2x+1)(x^2−2x+3)<3

Предыдущий
вопрос Следующий
вопрос

First, expand the left side of the inequality:

(x^2-2x+1)(x^2-2x+3) = x^4 - 2x^3 + 3x^2 - 2x^3 + 4x^2 - 6x + x^2 - 2x + 3
= x^4 - 4x^3 + 8x^2 - 8x + 3

Now we have to solve the inequality:

x^4 - 4x^3 + 8x^2 - 8x + 3 < 3

Subtract 3 from both sides:

x^4 - 4x^3 + 8x^2 - 8x < 0

Factor out x:

x(x^3 - 4x^2 + 8x - 8) < 0

Now, we need to find the critical points by setting each factor to 0:

x = 0

To determine the sign of the inequality, we can plug in test values for x:

For x < 0, we get a positive value
For x > 0, we get a negative value

Therefore, the solution to the inequality is:

x < 0

So, the solution to the inequality (x^2−2x+1)(x^2−2x+3) < 3 is x < 0.