(o) \( \frac{\sqrt{48}-\sqrt{32}}{\sqrt{12}-\sqrt{8}} \)
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To simplify the expression \( \frac{\sqrt{48}-\sqrt{32}}{\sqrt{12}-\sqrt{8}} \), let's first simplify the square roots. We know that: - \( \sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3} \) - \( \sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2} \) - \( \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} \) - \( \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2} \) Now substituting back into the expression: \[ \frac{4\sqrt{3} - 4\sqrt{2}}{2\sqrt{3} - 2\sqrt{2}} \] Factor out common terms in the numerator and denominator: \[ = \frac{4(\sqrt{3} - \sqrt{2})}{2(\sqrt{3} - \sqrt{2})} \] Now we can simplify this, cancelling the \((\sqrt{3} - \sqrt{2})\) terms (as long as \( \sqrt{3} \neq \sqrt{2}\)): \[ = \frac{4}{2} = 2 \] So, the simplified value of the expression \( \frac{\sqrt{48}-\sqrt{32}}{\sqrt{12}-\sqrt{8}} \) is \( 2 \).