$$\frac{ \sqrt{ 3+2 } }{ \sqrt{ 3-2 } } + \frac{ \sqrt{ 3-2 } }{ \sqrt{ 3+2 } } =a \sqrt{ 3 } +b$$
$a=-\frac{\sqrt{3}b}{3}+\frac{2\sqrt{15}}{5}$
$$\frac{\sqrt{5}}{\sqrt{3-2}}+\frac{\sqrt{3-2}}{\sqrt{3+2}}=a\sqrt{3}+b$$
$$\frac{\sqrt{5}}{\sqrt{1}}+\frac{\sqrt{3-2}}{\sqrt{3+2}}=a\sqrt{3}+b$$
$$\frac{\sqrt{5}}{1}+\frac{\sqrt{3-2}}{\sqrt{3+2}}=a\sqrt{3}+b$$
$$\sqrt{5}+\frac{\sqrt{3-2}}{\sqrt{3+2}}=a\sqrt{3}+b$$
$$\sqrt{5}+\frac{\sqrt{1}}{\sqrt{3+2}}=a\sqrt{3}+b$$
$$\sqrt{5}+\frac{1}{\sqrt{3+2}}=a\sqrt{3}+b$$
$$\sqrt{5}+\frac{1}{\sqrt{5}}=a\sqrt{3}+b$$
$$\sqrt{5}+\frac{\sqrt{5}}{\left(\sqrt{5}\right)^{2}}=a\sqrt{3}+b$$
$$\sqrt{5}+\frac{\sqrt{5}}{5}=a\sqrt{3}+b$$
$$\frac{6}{5}\sqrt{5}=a\sqrt{3}+b$$
$$a\sqrt{3}+b=\frac{6}{5}\sqrt{5}$$
$$a\sqrt{3}=\frac{6}{5}\sqrt{5}-b$$
$$\sqrt{3}a=-b+\frac{6\sqrt{5}}{5}$$
$$\frac{\sqrt{3}a}{\sqrt{3}}=\frac{-b+\frac{6\sqrt{5}}{5}}{\sqrt{3}}$$
$$a=\frac{-b+\frac{6\sqrt{5}}{5}}{\sqrt{3}}$$
$$a=\frac{\sqrt{3}\left(-5b+6\sqrt{5}\right)}{15}$$
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$b=-\sqrt{3}a+\frac{6\sqrt{5}}{5}$
$$b=\frac{6}{5}\sqrt{5}-a\sqrt{3}$$
$$b=-\sqrt{3}a+\frac{6}{5}\sqrt{5}$$
