Question

Please finish these sums alone

Answer 1

$\huge \mathfrak{answer \: 18}$

$\frac{ 6-4 \sqrt{ 3 } }{ 6+4 \sqrt{ 3 } } \\ \\\frac{\left(6-4\sqrt{3}\right)\left(6-4\sqrt{3}\right)}{\left(6+4\sqrt{3}\right)\left(6-4\sqrt{3}\right)} \\ \\\frac{\left(6-4\sqrt{3}\right)\left(6-4\sqrt{3}\right)}{6^{2}-\left(4\sqrt{3}\right)^{2}} \\ \\\frac{\left(6-4\sqrt{3}\right)^{2}}{6^{2}-\left(4\sqrt{3}\right)^{2}} \\ \\ \frac{36-48\sqrt{3}+16\left(\sqrt{3}\right)^{2}}{6^{2}-\left(4\sqrt{3}\right)^{2}} \\ \\\frac{36-48\sqrt{3}+16\times 3}{6^{2}-\left(4\sqrt{3}\right)^{2}} \\ \\ \frac{84-48\sqrt{3}}{6^{2}-\left(4\sqrt{3}\right)^{2}} \\ \\ \\\frac{84-48\sqrt{3}}{36-\left(4\sqrt{3}\right)^{2}} \\ \\\frac{84-48\sqrt{3}}{36-4^{2}\left(\sqrt{3}\right)^{2}} \\ \\ \frac{84-48\sqrt{3}}{36-16\left(\sqrt{3}\right)^{2}} \\ \\\frac{84-48\sqrt{3}}{36-16\times 3} \\ \\\frac{84-48\sqrt{3}}{36-48} \\ \\\frac{84-48\sqrt{3}}{-12} \\ \\ -7+4\sqrt{3}$

$\huge \mathfrak{answer \: 19}$

$\frac{ 2 \sqrt{ 3 } }{ \sqrt{ 3 } - \sqrt{ 2 } } + \frac{ 3 \sqrt{ 2 } }{ \sqrt{ 3 } + \sqrt{ 2 } }\\ \\\frac{2\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+\frac{3\sqrt{2}}{\sqrt{3}+\sqrt{2}} \\ \\\frac{2\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}+\frac{3\sqrt{2}}{\sqrt{3}+\sqrt{2}} \\ \\\frac{2\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}{3-2}+\frac{3\sqrt{2}}{\sqrt{3}+\sqrt{2}} \\ \\\frac{2\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}{1}+\frac{3\sqrt{2}}{\sqrt{3}+\sqrt{2}} \\ \\2\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)+\frac{3\sqrt{2}}{\sqrt{3}+\sqrt{2}} \\ \\2\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)+\frac{3\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)} \\ \\2\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)+\frac{3\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)}{\left(\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}} \\ \\2\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)+\frac{3\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)}{3-2} \\ \\2\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)+\frac{3\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)}{1} \\ \\2\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)+3\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right) \\ \\2\left(\sqrt{3}\right)^{2}+2\sqrt{3}\sqrt{2}+3\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right) \\ \\2\times 3+2\sqrt{3}\sqrt{2}+3\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right) \\ \\6+2\sqrt{3}\sqrt{2}+3\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right) \\ \\6+2\sqrt{6}+3\sqrt{2}\sqrt{3}-3\left(\sqrt{2}\right)^{2} \\ \\6+2\sqrt{6}+3\sqrt{6}-6 \\ \\6+5\sqrt{6}-6 \\ \\5\sqrt{6}$

$\huge \mathfrak{answer \: 18}$

\frac{ 7+3 \sqrt{ 5 } }{ 3+ \sqrt{ 5 } } + \frac{ 7-3 \sqrt{ 5 } }{ 3- \sqrt{ 5 } } =a+ \sqrt{ 5 } b \\ \\\frac{\left(7+3\sqrt{5}\right)\left(3-\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}+\frac{7-3\sqrt{5}}{3-\sqrt{5}}=a+\sqrt{5}b \\ \\ \frac{\left(7+3\sqrt{5}\right)\left(3-\sqrt{5}\right)}{3^{2}-\left(\sqrt{5}\right)^{2}}+\frac{7-3\sqrt{5}}{3-\sqrt{5}}=a+\sqrt{5}b\\ \\ \frac{\left(7+3\sqrt{5}\right)\left(3-\sqrt{5}\right)}{9-5}+\frac{7-3\sqrt{5}}{3-\sqrt{5}}=a+\sqrt{5}b\\ \\ \frac{\left(7+3\sqrt{5}\right)\left(3-\sqrt{5}\right)}{4}+\frac{7-3\sqrt{5}}{3-\sqrt{5}}=a+\sqrt{5}b\\ \\ \frac{\left(7+3\sqrt{5}\right)\left(3-\sqrt{5}\right)}{4}+\frac{\left(7-3\sqrt{5}\right)\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}=a+\sqrt{5}b \\ \\ \frac{\left(7+3\sqrt{5}\right)\left(3-\sqrt{5}\right)}{4}+\frac{\left(7-3\sqrt{5}\right)\left(3+\sqrt{5}\right)}{3^{2}-\left(\sqrt{5}\right)^{2}}=a+\sqrt{5}b\\ \\ \frac{\left(7+3\sqrt{5}\right)\left(3-\sqrt{5}\right)}{4}+\frac{\left(7-3\sqrt{5}\right)\left(3+\sqrt{5}\right)}{9-5}=a+\sqrt{5}b\\ \\ \frac{\left(7+3\sqrt{5}\right)\left(3-\sqrt{5}\right)}{4}+\frac{\left(7-3\sqrt{5}\right)\left(3+\sqrt{5}\right)}{4}=a+\sqrt{5}b\\ \\ \frac{\left(7+3\sqrt{5}\right)\left(3-\sqrt{5}\right)+\left(7-3\sqrt{5}\right)\left(3+\sqrt{5}\right)}{4}=a+\sqrt{5}b\\ \\ \frac{21-7\sqrt{5}+9\sqrt{5}-15+21+7\sqrt{5}-9\sqrt{5}-15}{4}=a+\sqrt{5}b \\ \\\frac{12}{4}=a+\sqrt{5}b 3 \\ =a+\sqrt{5}b a+\sqrt{5}b=3\\ \\ a=3-\sqrt{5}b\\ \\ a=-\sqrt{5}b+3