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Simplify:
$\Bigg ( \dfrac { \sqrt{ 6 } - \sqrt{ 5 } } { \sqrt{ 6 } + \sqrt{ 5 } } + \dfrac { \sqrt{ 6 } + \sqrt{ 5 } } { \sqrt{ 6 } - \sqrt{ 5 } } \Bigg )$

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Answer:

$22$

Step by Step Explanation:

1. Let us multiply the numerator and denominator of $\dfrac { \sqrt{ 6 } - \sqrt{ 5 } } { \sqrt{ 6 } + \sqrt{ 5 } }$ by its rationalising factor. $$\begin{align} & \dfrac { (\sqrt{ 6 } - \sqrt{ 5 }) } { (\sqrt{ 6 } + \sqrt{ 5 }) } \times \dfrac { (\sqrt{ 6 } - \sqrt{ 5 }) } { (\sqrt{ 6 } - \sqrt{ 5 }) } \\ = \space & \dfrac { (\sqrt{ 6 } - \sqrt{ 5 })^2 } { (\sqrt{ 6 } + \sqrt{ 5 })(\sqrt{ 6 } - \sqrt{ 5 }) } \\ = \space & \dfrac { (\sqrt{ 6 })^2 + (\sqrt{ 5 })^2 - 2 \times \sqrt{ 6 } \times \sqrt{ 5 } } { (\sqrt{ 6 })^2 - (\sqrt{ 5 })^2 } && [(a - b)^2 = a^2 + b^2 - 2ab \text{ and } (a + b)(a - b) = a^2 - b^2]\\ = \space & \dfrac { 6 + 5 - 2 \sqrt{ 30 } } { 6 - 5 } \\ = \space & 11 - 2 \sqrt{ 30 } \end{align}$$
2. Similarly, let us multiply the numerator and denominator of $\dfrac { (\sqrt{ 6 } + \sqrt{ 5 }) } { (\sqrt{ 6 } - \sqrt{ 5 }) }$ by its rationalising factor. $$\begin{align} & \dfrac { (\sqrt{ 6 } + \sqrt{ 5 }) } { (\sqrt{ 6 } - \sqrt{ 5 }) } \times \dfrac { (\sqrt{ 6 } + \sqrt{ 5 }) } { (\sqrt{ 6 } + \sqrt{ 5 }) } \\ = \space & \dfrac { (\sqrt{ 6 } + \sqrt{ 5 })^2 } { (\sqrt{ 6 } - \sqrt{ 5 })(\sqrt{ 6 } + \sqrt{ 5 }) } \\ = \space & \dfrac { (\sqrt{ 6 })^2 + (\sqrt{ 5 })^2 + 2 \times \sqrt{ 6 } \times \sqrt{ 5 } } { (\sqrt{ 6 })^2 - (\sqrt{ 5 })^2 } && [(a + b)^2 = a^2 + b^2 + 2ab \text{ and } (a + b)(a - b) = a^2 - b^2] \\ = \space & \dfrac { 6 + 5 + 2 \sqrt{ 30 } } { 6 - 5 } \\ = \space & 11 + 2 \sqrt{ 30 } \\ \end{align}$$
3. Thus, $$\begin{align} \Bigg ( \dfrac { \sqrt{ 6 } - \sqrt{ 5 } } { \sqrt{ 6 } + \sqrt{ 5 } } + \dfrac { \sqrt{ 6 } + \sqrt{ 5 } } { \sqrt{ 6 } - \sqrt{ 5 } } \Bigg ) = 11 - 2 \sqrt{ 30 } + 11 + 2 \sqrt{ 30 } = 22 \end{align}$$

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