6+25−(12)2+(1)2+2⋅12⋅1=6+25−(12+1)2=6+25−12−1=6+24−23=6+2⋅(3)2+(1)2−2⋅1⋅3=6+2(3−1)2=6+2(3−1)=6+23−2 ⇒4+23⇒(3)2+1)2+23.1⇒(3+1)2⇒3+1

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$\sqrt{6 + 2 \sqrt{5 - \sqrt{(13 + \sqrt{48})}}} =$

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$\sqrt{5} + 1$

$\sqrt{3} + 1$

$\sqrt{6} + \sqrt{2}$

$\sqrt{2} + 1$

answer is C.

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Detailed Solution

$\begin{matrix} \sqrt{6 + 2 \sqrt{5 - \sqrt{(\sqrt{12})^{2} + (1)^{2} + 2 \cdot \sqrt{12} \cdot 1}}} \\ = \sqrt{6 + 2 \sqrt{5 - \sqrt{{(\sqrt{12} + 1)}^{2}}}} \\ = \sqrt{6 + 2 \sqrt{5 - \sqrt{12} - 1}} \\ = \sqrt{6 + 2 \sqrt{4 - 2 \sqrt{3}}} \\ = \sqrt{6 + 2 \cdot \sqrt{(\sqrt{3})^{2} + (1)^{2} - 2 \cdot 1 \cdot \sqrt{3}}} \end{matrix}$
$\begin{matrix} = \sqrt{6 + 2 \sqrt{{(\sqrt{3} - 1)}^{2}}} \\ = \sqrt{6 + 2 (\sqrt{3} - 1)} \\ = \sqrt{6 + 2 \sqrt{3} - 2} \Rightarrow \sqrt{4 + 2 \sqrt{3}} \end{matrix}$
$\begin{array}{l} \Rightarrow \sqrt{(\sqrt{3})^{2} + (1)^{2} + 2 \sqrt{3} . 1} \\ \Rightarrow \sqrt{(\sqrt{3} + 1)^{2}} \\ \Rightarrow \sqrt{3} + 1 \end{array}$