(e) (1) Simplify \( \sqrt{b+\sqrt{2 b-1} \cdot \sqrt{b-\sqrt{2 b-1} \text { where } b>1 \text {. }}} \begin{array}{l}\text { (2) Without using a calculator, express }(\sqrt{6}-\sqrt{12})^{2} \text { in the form } \\ a(b+c \sqrt{2}) \text { and hence state the value of } a, b \text { and } c \text {. } \\ \text { (3) If } a \text { and } b \text { are natural numbers such that } \frac{14}{\sqrt{63}-\sqrt{28}}=a \sqrt{b} \text {, } \\ \text { determine, without using a calculator, the value of } a \text { and } b \text {. } \\ \text { (4) Express } \frac{2}{\sqrt{3}}-\frac{\sqrt{3}}{2} \text { in the form } \frac{\sqrt{a}}{b} \text { where } b \neq 0 \text { and } \sqrt{a} \text { is a surd. }\end{array} \) (G11075

Let's solve each part of the problem step by step. ### Part (1) We need to simplify the expression \( \sqrt{b+\sqrt{2 b-1} \cdot \sqrt{b-\sqrt{2 b-1}}} \). 1. **Simplify the inner expression**: \[ \sqrt{2b-1} \cdot \sqrt{b-\sqrt{2b-1}} = \sqrt{(2b-1)(b-\sqrt{2b-1})} \] Let's denote \( x = \sqrt{2b-1} \). Then, we have: \[ \sqrt{b-x} \] Now, we can rewrite the expression: \[ \sqrt{b + x \cdot \sqrt{b-x}} \] 2. **Substituting back**: We can express \( b + \sqrt{(2b-1)(b-\sqrt{2b-1})} \) in a simpler form. However, let's calculate it directly: \[ \sqrt{b + \sqrt{(2b-1)(b-\sqrt{2b-1})}} \] 3. **Final simplification**: After some algebraic manipulation, we find that: \[ \sqrt{b+\sqrt{(2b-1)(b-\sqrt{2b-1})}} = \sqrt{b + \sqrt{b^2 - 2b + 1}} = \sqrt{b + (b-1)} = \sqrt{2b-1} \] Thus, the simplified form is: \[ \sqrt{2b-1} \] ### Part (2) We need to express \( (\sqrt{6}-\sqrt{12})^{2} \) in the form \( a(b+c \sqrt{2}) \). 1. **Calculate \( (\sqrt{6}-\sqrt{12})^{2} \)**: \[ (\sqrt{6}-\sqrt{12})^{2} = 6 - 2\sqrt{6 \cdot 12} + 12 = 6 - 2\sqrt{72} + 12 = 18 - 2\sqrt{72} \] 2. **Simplify \( \sqrt{72} \)**: \[ \sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2} \] Thus, \[ 18 - 2(6\sqrt{2}) = 18 - 12\sqrt{2} \] 3. **Express in the required form**: \[ 18 - 12\sqrt{2} = 6(3 - 2\sqrt{2}) \] Here, \( a = 6 \), \( b = 3 \), and \( c = -2 \). ### Part (3) We need to determine \( a \) and \( b \) such that: \[ \frac{14}{\sqrt{63}-\sqrt{28}}=a \sqrt{b} \] 1. **Rationalize the denominator**: \[ \frac{14(\sqrt{63}+\sqrt{28})}{(\sqrt{63}-\sqrt{28})(\sqrt{63}+\sqrt{28})} = \frac{14(\sqrt{63}+\sqrt{28})}{63 - 28} = \frac{14(\sqrt{63}+\sqrt{28})}{35} \] Simplifying gives: \[ \frac{2(\sqrt{63}+\sqrt{28})}{5} \] 2. **Express in the form \( a \sqrt{b} \)**: \[ \frac{2}{5}(\sqrt{63}+\sqrt{28}) = \frac{2}{5}(\sqrt{63} + 2\sqrt{7}) \] Here, \( a = \frac{2}{5} \) and \( b = 63 \) or \( b = 28 \). ### Part (4) We need to express \( \frac{2}{\sqrt{3}}-\frac{\sqrt{3}}{2} \) in the form \( \frac{\sqrt{a}}{b} \). 1. **Find a common denominator**: \[ \frac{2}{\sqrt{3}} - \frac{\sqrt{3}}{2} = \frac{4 - 3}{2\sqrt{3}} = \frac{1}{2\sqrt{3}} \] 2. **Rationalize the denominator**: \[ \frac{1}{2\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{6} \] Thus, we have \( a = 3 \) and \( b = 6 \). ### Summary of Results 1. Part (1): \( \sqrt{2b-1} \) 2. Part (2): \( a = 6, b = 3, c = -2 \) 3. Part (3): \( a = 2, b = 63 \) or \( b = 28 \) 4. Part (4): \( a = 3, b = 6 \)