Question 1: \((iv)\) \(\quad\)\(\large{\frac{6-4\sqrt{2}}{6+4\sqrt{2}}} \)
Solution:
\[\begin{align*}
\frac{6-4\sqrt{2}}{6+4\sqrt{2}}&=\frac{2\left( 3-2\sqrt{2}\right)}{2\left( 3+2\sqrt{2}\right)}\\
&=\frac{3-2\sqrt{2}}{3+2\sqrt{2}} \times \frac{3-2\sqrt{2}}{3-2\sqrt{2}}\\
&=\frac{\left( 3-2\sqrt{2}\right)^{2}}{3^{2} -\left( 2\sqrt{2}\right)^{2}}\\
&=\small{\frac{3^{2} +\left( 2\sqrt{2}\right)^{2} +2\times 3\times 2\sqrt{2}}{9-4\times 2}}\\
&=\frac{9+4\times 2-12\sqrt{2}}{9-8}\\
&=9+8-12\sqrt{2}\\
&=17-12\sqrt{2}\\
\end{align*}\]
——– Alternate Method ——–
\[\begin{align*}
\frac{6-4\sqrt{2}}{6+4\sqrt{2}}&=\frac{6-4\sqrt{2}}{6+4\sqrt{2}} \times \frac{6-4\sqrt{2}}{6-4\sqrt{2}}\\
&=\frac{\left( 6-4\sqrt{2}\right)^{2}}{6^{2} -\left( 4\sqrt{2}\right)^{2}}\\
&=\small{\frac{6^{2} +\left( 4\sqrt{2}\right)^{2} -2\times 6\times 4\sqrt{2}}{36-16\times 2}}\\
&=\frac{36+16\times 2-48\sqrt{2}}{36-32}\\
&=\frac{36+32-48\sqrt{2}}{4}\\
&=\frac{68-48\sqrt{2}}{4}\\
& = \frac{4\left( 17-12\sqrt{2}\right)}{4}\\
&=17-12\sqrt{2}\\
\end{align*}\]