Question

Please solve 8th and 9th question
No spamming please

Answer 1

hey there❤

refer to attachment

Answer 2

Answer:

Option(2)

Step-by-step explanation:

In continuation to Dhruv15819......

(8)

$Given:\frac{2(\sqrt{2}+\sqrt{6})}{3\sqrt{2+\sqrt{3}}}+\sqrt{2+\sqrt{3}} + \sqrt{2-\sqrt{3}}$

$\Longrightarrow \frac{2(\sqrt{2}+\sqrt{6})}{3\sqrt{\frac{4+2\sqrt{3}}{2}}} + \sqrt{\frac{2(2+\sqrt{3})}{2}}+\sqrt{\frac{2(2-\sqrt{3})}{2}}$

$\Longrightarrow{\frac{2(\sqrt{2}+\sqrt{6})}{\sqrt\frac{3(3+1+2\sqrt{3})}{2}}} + \sqrt{\frac{4+2\sqrt{3} }{2}}+\sqrt{\frac{4-2\sqrt{3} }{2}}$

$\Longrightarrow{\frac{2(\sqrt{2}+\sqrt{6})}{\sqrt\frac{(3\sqrt{3}+1)^2}{2}}}+\sqrt{\frac{3+1+2\sqrt{3}}{2}}+\sqrt{\frac{3+1-2\sqrt{3}}{2}}$

$\Longrightarrow{\frac{2(\sqrt{2}+\sqrt{6})}{\sqrt\frac{(3\sqrt{3}+1)^2}{2}}}+\sqrt{\frac{(\sqrt{3})^2+1+2\sqrt{3}}{2}}+\sqrt{\frac{(\sqrt{3})^2+1-2\sqrt{3}}{2}}$

$\Longrightarrow{\frac{2(\sqrt{2}+\sqrt{6})}{\frac{3(\sqrt{3}+1)}{\sqrt2}}}+\sqrt{\frac{(\sqrt{3}+1)^2}{2}}+\sqrt{\frac{(\sqrt{3}-1)^2}{2}}$

$\Longrightarrow\frac{2\sqrt{2}(1+\sqrt{3})}{\frac{3(\sqrt{3}+1)}{\sqrt{2}}}+\frac{\sqrt{3}+1}{\sqrt{2}}+\frac{\sqrt{3}-1}{\sqrt{2}}$

$\Longrightarrow\frac{2\sqrt{2}\sqrt{2}}{3}+\frac{\sqrt{3}+1}{\sqrt{2}}+\frac{\sqrt{3}-1}{\sqrt{2}}$

$\Longrightarrow\frac{4}{3}+\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt2}$

$\Longrightarrow\frac{4}{3}+\frac{2\sqrt{6}}{\sqrt{2}}$

$\Longrightarrow\frac{4}{3}+\sqrt{2}\sqrt{3}$

$\Longrightarrow\frac{4}{3}+\sqrt{6}$

$\Longrightarrow\boxed{\frac{4+3\sqrt{6}}{3}}$

Hope it helps!