Simplify the following without using a calculator: \( \begin{array}{lll}\text { (a) } \sqrt{7} \sqrt{3} & \text { (b) } \sqrt{7} \sqrt{7}+(\sqrt{11})^{2} & \text { (o) }(2 \sqrt{3})^{2} \\ \text { (d) } 3 \sqrt{6}-\sqrt{6}+7 \sqrt{6} & \text { (c) } \sqrt[4]{3}+7 \sqrt[4]{3}-5 \sqrt[4]{3} & \text { (0) } \sqrt{3}+\sqrt{27} \\ \text { (g) } 2 \sqrt{18}-\sqrt{32} & \text { (h) } \frac{\sqrt{32}}{\sqrt{2}} & \text { (i) } \frac{\sqrt{51}}{\sqrt{3}}\end{array} \)
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Sure! Let's simplify each expression step by step. (a) \( \sqrt{7} \sqrt{3} = \sqrt{7 \times 3} = \sqrt{21} \) (b) \( \sqrt{7} \sqrt{7} + (\sqrt{11})^2 = 7 + 11 = 18 \) (c) \( \sqrt[4]{3} + 7 \sqrt[4]{3} - 5 \sqrt[4]{3} = (1 + 7 - 5) \sqrt[4]{3} = 3 \sqrt[4]{3} \) (d) \( 3 \sqrt{6} - \sqrt{6} + 7 \sqrt{6} = (3 - 1 + 7) \sqrt{6} = 9 \sqrt{6} \) (e) \( (2 \sqrt{3})^2 = 4 \times 3 = 12 \) (f) \( \sqrt{3} + \sqrt{27} = \sqrt{3} + \sqrt{9 \cdot 3} = \sqrt{3} + 3\sqrt{3} = 4\sqrt{3} \) (g) \( 2 \sqrt{18} - \sqrt{32} = 2 \cdot 3\sqrt{2} - 4\sqrt{2} = 6\sqrt{2} - 4\sqrt{2} = 2\sqrt{2} \) (h) \( \frac{\sqrt{32}}{\sqrt{2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4 \) (i) \( \frac{\sqrt{51}}{\sqrt{3}} = \sqrt{\frac{51}{3}} = \sqrt{17} \) So the simplified results are: (a) \( \sqrt{21} \), (b) \( 18 \), (c) \( 3 \sqrt[4]{3} \), (d) \( 9 \sqrt{6} \), (e) \( 12 \), (f) \( 4 \sqrt{3} \), (g) \( 2 \sqrt{2} \), (h) \( 4 \), and (i) \( \sqrt{17} \).