\(\sqrt{13+\sqrt{48}}=\sqrt{13+\sqrt{4.12}}=\sqrt{13+2\sqrt{12}}=\sqrt{\left(\sqrt{12}+1\right)^2}\)

\(=\sqrt{12}+1=2\sqrt{3}+1\)

\(\Rightarrow\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{5-2\sqrt{3}-1}=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(=\sqrt{3}-1\)

\(\Rightarrow\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{3+\sqrt{3}-1}=\sqrt{2+\sqrt{3}}\)

\(\Rightarrow\sqrt{\dfrac{4+2\sqrt{3}}{2}}=\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{2}}=\dfrac{\sqrt{3}+1}{\sqrt{2}}\)

\(\Rightarrow2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}==2.\dfrac{\sqrt{3}+1}{\sqrt{2}}=\sqrt{6}+\sqrt{2}\)

2) biến đổi khúc sau như câu 1:

\(\Rightarrow\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}\)

\(=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)

4) Ta có: \(\sqrt{30-2\sqrt{16+6\sqrt{11+4\sqrt{4-2\sqrt{3}}}}}\)

\(=\sqrt{30-2\sqrt{16+6\sqrt{11+4\left(\sqrt{3}-1\right)}}}\)

\(=\sqrt{30-2\sqrt{16+6\sqrt{7+4\sqrt{3}}}}\)

\(=\sqrt{30-2\sqrt{16+6\left(2+\sqrt{3}\right)}}\)

\(=\sqrt{30-2\sqrt{28+6\sqrt{3}}}\)

\(=\sqrt{30-2\left(3\sqrt{3}+1\right)}\)

\(=\sqrt{28-6\sqrt{3}}=3\sqrt{3}-1\)

5) Ta có: \(\dfrac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)

\(=\dfrac{5\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)^2}{\sqrt{75}-5\sqrt{2}}\)

\(=\dfrac{5\left(\sqrt{3}-\sqrt{2}\right)}{5\left(\sqrt{3}-\sqrt{2}\right)}=1\)

Ta có : \(2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}\) \(=2\sqrt{3+\sqrt{5-\sqrt{12+2\sqrt{12}+1}}}\)

\(=2\sqrt{3+\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}\)\(=2\sqrt{3+\sqrt{5-\sqrt{12}-1}}\)

\(=2\sqrt{3+\sqrt{3-2\sqrt{3}+1}}\)\(=2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}=2\sqrt{3+\sqrt{3}-1}\)

\(=2\sqrt{2+\sqrt{3}}=\sqrt{2}\sqrt{4+2\sqrt{3}}=\sqrt{2}\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{2}\left(\sqrt{3}+1\right)\)

Ta có : \(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)\(=\sqrt{6+2\sqrt{5-\sqrt{12+2\sqrt{12}+1}}}\)

\(=\sqrt{6+2\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}\)\(=\sqrt{6+2\sqrt{4-\sqrt{12}}}=\sqrt{6+2\sqrt{3-2\sqrt{3}+1}}\)

\(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{6+2\sqrt{3}-2}\)

\(=\sqrt{4+2\sqrt{3}}=\sqrt{3+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)

\(A=\sqrt{3+\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}=\sqrt{3+\sqrt{5-\left(2\sqrt{3}+1\right)}}\)

\(=\sqrt{3+\sqrt{4-2\sqrt{3}}}=\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}\)

\(=\sqrt{3+\sqrt{3}-1}=\sqrt{2+\sqrt{3}}=\dfrac{1}{\sqrt{2}}\sqrt{4+2\sqrt{3}}\)

\(=\dfrac{1}{\sqrt{2}}\sqrt{\left(\sqrt{3}+1\right)^2}=\dfrac{\sqrt{3}+1}{\sqrt{2}}=\dfrac{\sqrt{2}+\sqrt{6}}{2}\)

\(A=\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}\\ =\sqrt{3+\sqrt{5-\sqrt{\left(1+2\sqrt{3}\right)^2}}}\\ =\sqrt{3+\sqrt{5-1+2\sqrt{3}}}\\ =\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}\\ =\sqrt{3+\sqrt{3}-1}\\ =\sqrt{2+\sqrt{3}}\)

Ta có: \(A=\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}\)

\(=\sqrt{3+\sqrt{4-2\sqrt{3}}}\)

\(=\sqrt{3+\sqrt{3}-1}\)

\(=\sqrt{2-\sqrt{3}}\)

\(=\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{6}-\sqrt{2}}{2}\)

\(=\dfrac{2\cdot\sqrt{3+\sqrt{5}-2\sqrt{3}-1}}{\sqrt{6}+\sqrt{2}}=\dfrac{2\cdot\sqrt{2-2\sqrt{3}+\sqrt{5}}}{\sqrt{6}+\sqrt{2}}\)

Có:\(\sqrt{48}< \sqrt{49}=7\)

\(13-\sqrt{35}>13-\sqrt{36}=7\)

\(\Rightarrow\sqrt{48}< 13-\sqrt{35}\)

\(\sqrt{48}+\sqrt{35}< \sqrt{49}+\sqrt{36}=7+6=13\)

\(\rightarrow\sqrt{48}< 13-\sqrt{35}\)

Ta có:

\(\sqrt{48}< \sqrt{49}=7\left(1\right)\)

\(13-\sqrt{35}>13-\sqrt{36}=13-6=7\left(2\right)\)

Từ (1) và (2) :

Suy ra: \(\sqrt{48}< 13-\sqrt{35}\)

\(\)

Lời giải:

\(\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{5-\sqrt{13+2\sqrt{12}}}=\sqrt{5-\sqrt{12+2\sqrt{12}+1}}\)

\(=\sqrt{5-\sqrt{(\sqrt{12}+1)^2}}=\sqrt{5-(\sqrt{12}+1)}=\sqrt{4-\sqrt{12}}\)

\(=\sqrt{4-2\sqrt{3}}=\sqrt{3-2\sqrt{3}+1}=\sqrt{(\sqrt{3}-1)^2}=\sqrt{3}-1\)