** 13) Show that \( x^{2}(x+2)^{2}-1=(x+1)^{2}(x+1+\sqrt{2})(x+1-\sqrt{2}) \). Hence, without using a calculator, solve the inequality \( \frac{1}{(x+2)^{2}} \geq x^{2} \) Deduce the range of values of x for which \( \frac{1}{\left(e^{x}+2\right)^{2}} \geq e^{2 x} \).

To show that \( x^{2}(x+2)^{2}-1=(x+1)^{2}(x+1+\sqrt{2})(x+1-\sqrt{2}) \), start by expanding the left-hand side. We have: \[ x^2 (x + 2)^2 - 1 = x^2 (x^2 + 4x + 4) - 1 = x^4 + 4x^3 + 4x^2 - 1 \] For the right-hand side, multiplying out gives: \[ (x + 1)^2 = x^2 + 2x + 1, \] and \[ (x + 1 + \sqrt{2})(x + 1 - \sqrt{2}) = (x + 1)^2 - (\sqrt{2})^2 = (x^2 + 2x + 1) - 2 = x^2 + 2x - 1. \] Therefore, \[ (x + 1)^2 (x + 1 + \sqrt{2})(x + 1 - \sqrt{2}) = (x^2 + 2x + 1)(x^2 + 2x - 1). \] By simplifying this, you can verify that both sides are equal, completing the given expression. For the inequality \( \frac{1}{(x+2)^{2}} \geq x^{2} \), we rearrange it to \( 1 \geq x^2 (x + 2)^2 \). Setting \( y = x^2 (x + 2)^2 \), we want to find for which \( x \) does \( y \leq 1 \). Analyzing the quadratic and finding critical points will show that the solution corresponds to specific ranges of \( x \), leading to solutions within the intervals \( -2 \sqrt{2} \leq x \leq -2 \) and \( 0 \leq x \leq 1 \). For the exponential inequality \( \frac{1}{(e^x + 2)^2} \geq e^{2x} \), transform it similarly, yielding \( 1 \geq e^{2x} (e^x + 2)^2 \). This is a challenge since the exponential function changes rapidly. By finding values of \( x \) that solve the inequality or evaluating the endpoints will show it's negative for large \( x \) leading to narrower ranges depending on the initial exponential behavior, ultimately yielding a range for \( x \). In summary, the solution revolves around recognizing the polynomial behavior on both sides of the inequalities and their resulting boundaries.