Show that: \( \frac{2}{1+\sqrt{2}}-\frac{8}{\sqrt{8}}=-2 \)

To show that \( \frac{2}{1+\sqrt{2}}-\frac{8}{\sqrt{8}}=-2 \), we can simplify each term on the left side and then check if the expression equals \(-2\). First, simplify \( \frac{8}{\sqrt{8}} \): \[ \sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2} \quad \Rightarrow \quad \frac{8}{\sqrt{8}} = \frac{8}{2\sqrt{2}} = \frac{4}{\sqrt{2}}. \] Next, write our equation with this simplification: \[ \frac{2}{1+\sqrt{2}} - \frac{4}{\sqrt{2}}. \] Now, we need to simplify \( \frac{2}{1+\sqrt{2}} \). To do this, let's rationalize the denominator: \[ \frac{2}{1+\sqrt{2}} \cdot \frac{1-\sqrt{2}}{1-\sqrt{2}} = \frac{2(1-\sqrt{2})}{(1+\sqrt{2})(1-\sqrt{2})} = \frac{2(1-\sqrt{2})}{1^2 - (\sqrt{2})^2} = \frac{2(1-\sqrt{2})}{1 - 2} = \frac{2(1-\sqrt{2})}{-1} = -2(1-\sqrt{2}) = -2 + 2\sqrt{2}. \] Now rewrite the original expression: \[ -2 + 2\sqrt{2} - \frac{4}{\sqrt{2}}. \] Next, simplify \( - \frac{4}{\sqrt{2}} \): \[ - \frac{4}{\sqrt{2}} = -4 \cdot \frac{\sqrt{2}}{2} = -2\sqrt{2}. \] Putting it all together: \[ -2 + 2\sqrt{2} - 2\sqrt{2} = -2 + (2\sqrt{2} - 2\sqrt{2}) = -2 + 0 = -2. \] Thus, we have shown that: \[ \frac{2}{1+\sqrt{2}} - \frac{8}{\sqrt{8}} = -2. \] Therefore, the statement is verified: \[ \frac{2}{1+\sqrt{2}} - \frac{8}{\sqrt{8}} = -2. \]