Simplify the following without using a calculator. \( \begin{array}{lll}\text { (a) } \sqrt{7} \sqrt{3} & \text { (b) } \sqrt{7} \sqrt{7}+(\sqrt{11})^{2} & \text { (c) }(2 \sqrt{3})^{2} \\ \text { (d) } 3 \sqrt{6}-\sqrt{6}+7 \sqrt{6} & \text { (c) } \sqrt[4]{3}+7 \sqrt[4]{3}-5 \sqrt[4]{3} & \text { (f) } \sqrt{3}+\sqrt{2} \\ \text { (d) } 2 \sqrt{18}-\sqrt{32} & \text { (h) } \frac{\sqrt{32}}{\sqrt{2}} & \text { (i) } \frac{\sqrt{51}}{2 \sqrt{3}}\end{array} \)

Let's simplify each expression step by step without using a calculator. ### (a) \( \sqrt{7} \cdot \sqrt{3} \) Use the property \( \sqrt{a} \cdot \sqrt{b} = \sqrt{ab} \): \[ \sqrt{7} \cdot \sqrt{3} = \sqrt{7 \times 3} = \sqrt{21} \] **Answer:** \( \sqrt{21} \) --- ### (b) \( \sqrt{7} \cdot \sqrt{7} + (\sqrt{11})^{2} \) Simplify each term: \[ \sqrt{7} \cdot \sqrt{7} = \sqrt{7^2} = 7 \] \[ (\sqrt{11})^{2} = 11 \] Add them: \[ 7 + 11 = 18 \] **Answer:** \( 18 \) --- ### (c) \( (2 \sqrt{3})^{2} \) Square the expression: \[ (2 \sqrt{3})^{2} = 2^2 \cdot (\sqrt{3})^2 = 4 \cdot 3 = 12 \] **Answer:** \( 12 \) --- ### (d) \( 3 \sqrt{6} - \sqrt{6} + 7 \sqrt{6} \) Combine like terms: \[ 3 \sqrt{6} - \sqrt{6} + 7 \sqrt{6} = (3 - 1 + 7) \sqrt{6} = 9 \sqrt{6} \] **Answer:** \( 9 \sqrt{6} \) --- ### (e) \( \sqrt[4]{3} + 7 \sqrt[4]{3} - 5 \sqrt[4]{3} \) Combine like terms: \[ \sqrt[4]{3} + 7 \sqrt[4]{3} - 5 \sqrt[4]{3} = (1 + 7 - 5) \sqrt[4]{3} = 3 \sqrt[4]{3} \] **Answer:** \( 3 \sqrt[4]{3} \) --- ### (f) \( \sqrt{3} + \sqrt{2} \) This expression is already in its simplest form as the terms contain different radicals. **Answer:** \( \sqrt{3} + \sqrt{2} \) --- ### (g) \( 2 \sqrt{18} - \sqrt{32} \) Simplify each radical: \[ \sqrt{18} = \sqrt{9 \times 2} = 3 \sqrt{2} \] \[ \sqrt{32} = \sqrt{16 \times 2} = 4 \sqrt{2} \] Substitute back: \[ 2 \sqrt{18} - \sqrt{32} = 2 \times 3 \sqrt{2} - 4 \sqrt{2} = 6 \sqrt{2} - 4 \sqrt{2} = 2 \sqrt{2} \] **Answer:** \( 2 \sqrt{2} \) --- ### (h) \( \frac{\sqrt{32}}{\sqrt{2}} \) Simplify the fraction: \[ \frac{\sqrt{32}}{\sqrt{2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4 \] **Answer:** \( 4 \) --- ### (i) \( \frac{\sqrt{51}}{2 \sqrt{3}} \) Rationalize the denominator: \[ \frac{\sqrt{51}}{2 \sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{51 \times 3}}{2 \times 3} = \frac{\sqrt{153}}{6} \] Simplify \( \sqrt{153} \): \[ \sqrt{153} = \sqrt{9 \times 17} = 3 \sqrt{17} \] Substitute back: \[ \frac{3 \sqrt{17}}{6} = \frac{\sqrt{17}}{2} \] **Answer:** \( \frac{\sqrt{17}}{2} \) --- All parts have been simplified as requested.