Question

koi ye question bta do please​

Answer 1

$\text{To find}\:\:\sqrt{3+\sqrt{3}+\sqrt{2+\sqrt{3}+\sqrt{7+\sqrt{48}}}}\\\\\\\sqrt{3+\sqrt{3}+\sqrt{2+\sqrt{3}+\sqrt{7+\sqrt{48}}}}\\\\\\=\sqrt{3+\sqrt{3}+\sqrt{2+\sqrt{3}+\sqrt{7+\sqrt{4\times12}}}}\\\\\\=\sqrt{3+\sqrt{3}+\sqrt{2+\sqrt{3}+\sqrt{(3+4)+2\sqrt{3\times4}}}}\\\\\\=\sqrt{3+\sqrt{3}+\sqrt{2+\sqrt{3}+\sqrt{2^2+(\sqrt3)^2+(2\times2\sqrt{3})}}}\\\\\\=\sqrt{3+\sqrt{3}+\sqrt{2+\sqrt{3}+\sqrt{(2+\sqrt{3})^2}}}}$

$=\sqrt{3+\sqrt{3}+\sqrt{4+2\sqrt{3}}}\\\\\\=\sqrt{3+\sqrt{3}+\sqrt{3+1+(2\times1\times\sqrt{3})}}\\\\\\=\sqrt{3+\sqrt{3}+\sqrt{(\sqrt{3})^2+(1)^2+(2\times1\times\sqrt{3})}}\\\\\\=\sqrt{3+\sqrt{3}+\sqrt{(\sqrt{3}+1)^2}}\\\\\\=\sqrt{3+\sqrt{3}+(\sqrt{3}+1)}\\\\\\=\sqrt{3+2\sqrt{3}+1}\\\\\\=\sqrt{(\sqrt{3})^2+(1)^2+(2\times1\times\sqrt{3})}\\\\\\=\sqrt{(\sqrt{3}+1)^2}\\\\\\=\sqrt{3}+1$