EXERCISE 5 Simplify the following without using a calculator: \( \begin{array}{lll}\text { (a) } \sqrt{7} \sqrt{3} & \text { (b) } \sqrt{7} \sqrt{7}+(\sqrt{11})^{2} & \text { (c) }(2 \sqrt{3})^{2} \\ \text { (d) } 3 \sqrt{6}-\sqrt{6}+7 \sqrt{6} & \text { (c) } \sqrt[4]{3}+7 \sqrt[4]{3}-5 \sqrt[4]{3} & \text { (f) } \sqrt{3}+\sqrt{27} \\ \text { (g) } 2 \sqrt{18}-\sqrt{32} & \text { (h) } \frac{\sqrt{32}}{\sqrt{2}} & \text { (i) } \frac{\sqrt{51}}{\sqrt{3}}\end{array} \)

Let's simplify each part step-by-step: ### (a) \( \sqrt{7} \cdot \sqrt{3} \) Using the property \( \sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b} \): \[ \sqrt{7} \cdot \sqrt{3} = \sqrt{21} \] **Answer:** \( \sqrt{21} \) --- ### (b) \( \sqrt{7} \cdot \sqrt{7} + (\sqrt{11})^{2} \) First, simplify each term: 1. \( \sqrt{7} \cdot \sqrt{7} = \sqrt{7^2} = 7 \) 2. \( (\sqrt{11})^{2} = 11 \) Adding them together: \[ 7 + 11 = 18 \] **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) \( \sqrt{3} + \sqrt{27} \) Simplify \( \sqrt{27} \): \[ \sqrt{27} = \sqrt{9 \cdot 3} = \sqrt{9} \cdot \sqrt{3} = 3 \sqrt{3} \] Add the terms: \[ \sqrt{3} + 3 \sqrt{3} = 4 \sqrt{3} \] **Answer:** \( 4 \sqrt{3} \) --- ### (g) \( 2 \sqrt{18} - \sqrt{32} \) Simplify each radical: 1. \( \sqrt{18} = \sqrt{9 \cdot 2} = 3 \sqrt{2} \) 2. \( \sqrt{32} = \sqrt{16 \cdot 2} = 4 \sqrt{2} \) Now substitute back: \[ 2 \sqrt{18} - \sqrt{32} = 2 \cdot 3 \sqrt{2} - 4 \sqrt{2} = 6 \sqrt{2} - 4 \sqrt{2} = 2 \sqrt{2} \] **Answer:** \( 2 \sqrt{2} \) --- ### (h) \( \frac{\sqrt{32}}{\sqrt{2}} \) Simplify the fraction under the square root: \[ \frac{\sqrt{32}}{\sqrt{2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4 \] **Answer:** \( 4 \) --- ### (i) \( \frac{\sqrt{51}}{\sqrt{3}} \) Simplify the fraction under the square root: \[ \frac{\sqrt{51}}{\sqrt{3}} = \sqrt{\frac{51}{3}} = \sqrt{17} \] **Answer:** \( \sqrt{17} \) --- ### Summary of Answers - **(a)** \( \sqrt{21} \) - **(b)** \( 18 \) - **(c)** \( 12 \) - **(d)** \( 9 \sqrt{6} \) - **(e)** \( 3 \sqrt[4]{3} \) - **(f)** \( 4 \sqrt{3} \) - **(g)** \( 2 \sqrt{2} \) - **(h)** \( 4 \) - **(i)** \( \sqrt{17} \)