Question

(x^2+2x)^2-2(x+1)^2-1<_0

Answer 1

Ответ:

$x = - 3$

$x = 1$

Answer 2

Let's simplify the expression step by step:

(x^2 + 2x)^2 - 2(x + 1)^2 - 1 < 0

Expanding the square terms:

(x^4 + 4x^3 + 4x^2) - 2(x^2 + 2x + 1) - 1 < 0

Simplifying further:

x^4 + 4x^3 + 4x^2 - 2x^2 - 4x - 2 - 1 < 0

Combine like terms:

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

Now, we need to solve this inequality. To do that, we can find the critical points by setting the expression equal to zero:

x^4 + 4x^3 + 2x^2 - 4x - 3 = 0

However, solving this quartic equation analytically can be quite complex. In general, there is no simple method to solve quartic equations. You might need to use numerical methods or approximation techniques to find the exact solution.

Once you have the critical points, you can test the intervals between them to determine when the expression is less than zero. However, since the equation involves a quartic polynomial, the inequality may have multiple intervals and require more advanced methods to solve accurately.

Alternatively, if you provide specific values for x or a specific range of x-values, I can help you determine if the inequality is satisfied within that range.