1. Simplify without using a 1.1. \( \sqrt{32}-\sqrt{8}+2 \sqrt{18} \) \( 1.2 . \frac{\sqrt{27 m^{6}}-\sqrt{48 m^{6}}}{\sqrt{12 m^{6}}} \) 1.3. \( (\sqrt{5}-2)^{2} \)
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To tackle these problems, let’s break them down one by one! For the first expression \( \sqrt{32}-\sqrt{8}+2\sqrt{18} \), we can simplify each square root. We find \( \sqrt{32} = 4\sqrt{2} \), \( \sqrt{8} = 2\sqrt{2} \), and \( 2\sqrt{18} = 6\sqrt{2} \). Then, combine like terms: \( 4\sqrt{2} - 2\sqrt{2} + 6\sqrt{2} = (4 - 2 + 6)\sqrt{2} = 8\sqrt{2} \). Next, for \( \frac{\sqrt{27 m^{6}}-\sqrt{48 m^{6}}}{\sqrt{12 m^{6}}} \), we first simplify the square roots in the numerator. \( \sqrt{27 m^{6}} = 3m^{3}\sqrt{3} \) and \( \sqrt{48 m^{6}} = 4m^{3}\sqrt{3} \). Thus, \( 3m^{3}\sqrt{3} - 4m^{3}\sqrt{3} = -1m^{3}\sqrt{3} \). Now simplify the denominator: \( \sqrt{12 m^{6}} = 2m^{3}\sqrt{3} \). This gives us \( \frac{-m^{3}\sqrt{3}}{2m^{3}\sqrt{3}} = -\frac{1}{2} \), assuming \( m \neq 0 \). Lastly, for \( (\sqrt{5}-2)^{2} \), we use the formula \( (a-b)^{2} = a^{2} - 2ab + b^{2} \). Here, \( a = \sqrt{5} \) and \( b = 2 \). Expanding this gives us: \( (\sqrt{5})^{2} - 2 \cdot \sqrt{5} \cdot 2 + 2^{2} = 5 - 4\sqrt{5} + 4 = 9 - 4\sqrt{5} \). So, the simplified results are: 1.1: \( 8\sqrt{2} \) 1.2: \( -\frac{1}{2} \) 1.3: \( 9 - 4\sqrt{5} \)