(-16) : (((х+2) ²)-5) ≥ 0

To solve the inequality \((-16) : (((x+2)²)-5) ≥ 0\), we start by rewriting it in a more manageable form. The expression \(-16\) divided by another expression is greater than or equal to zero.

First, let's break down the inequality:

\[ -\frac{16}{((x+2)² - 5)} \geq 0 \]

Multiplying both sides of the inequality by \(-1\) (remembering to flip the inequality sign) gives us:

\[ \frac{16}{((x+2)² - 5)} \leq 0 \]

Now, we need to determine when the fraction \(\frac{16}{((x+2)² - 5)}\) is less than or equal to zero. This fraction will be non-positive (i.e., less than or equal to zero) when the numerator is positive (which it always is since \(16\) is positive) and the denominator \(((x+2)² - 5)\) is negative or zero.

Now, let's find the values of \(x\) that make the denominator \(((x+2)² - 5)\) equal to zero:

\[ (x+2)² - 5 = 0 \]

Solving this gives:

\[ (x+2)² = 5 \]

Taking square roots:

\[ x + 2 = \pm \sqrt{5} \]

Thus:

\[ x = -2 + \sqrt{5} \quad \text{or} \quad x = -2 - \sqrt{5} \]

Now we need to check when \((x+2)² - 5 < 0\).

The critical points obtained are \(x = -2 + \sqrt{5}\) and \(x = -2 - \sqrt{5}\). To analyze the sign of the denominator in intervals defined by these critical points, we check intervals:

1. \(x < -2 - \sqrt{5}\)
2. \(-2 - \sqrt{5} < x < -2 + \sqrt{5}\)
3. \(x > -2 + \sqrt{5}\)

Checking intervals:

1. For \(x < -2 - \sqrt{5}\): Choose \(x = -10\): \((x+2) = -8\) so \((x + 2)^2 = 64 - 5 = 59\) (positive).

2. For \(-2 - \sqrt{5} < x < -2 + \sqrt{5}\): Choose \(x = -2\): \((x + 2) = 0\) so \((x + 2)^2 = 0 - 5 = -5\) (negative).

3. For \(x > -2 + \sqrt{5}\): Choose \(x = 2\): \((x + 2) = 4\) so \((x + 2)^2 = 16 - 5 = 11\) (positive).

Conclusion:

The denominator \((x + 2)^{2} - 5 \leq 0\) when \(-2 - \sqrt{5} < x < -2 + \sqrt{5}\).

At the endpoints:

• At \(x = -2 + \sqrt{5}\), the denominator equals \(0\).
• At \(x = -2 - \sqrt{5}\), the denominator also equals \(0\).

Thus, the solution set for the original inequality \((-16) : (((x+2)²)-5) ≥ 0\) is:

\[ x \in [-2 - \sqrt{5}, -2 + \sqrt{5}] \]

Or, in interval notation:

\[ [-2 - \sqrt{5}, -2 + \sqrt{5}] \]