a) \(=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{16-2.4\sqrt{2}+2}}}\)

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

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

b) Tương tự a) đ/s =5

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

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

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

\(B=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}\)

\(B=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)

\(B=\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}=\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)

\(B=\sqrt{25}=5\)

Ta có: \(B=\sqrt{13+3\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)

\(=\sqrt{13+3\sqrt{2+\sqrt{8+2\cdot2\sqrt{2}\cdot1+1}}}\)

\(=\sqrt{13+3\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\)

\(=\sqrt{13+3\sqrt{2+2\sqrt{2}+1}}\)

\(=\sqrt{13+3\sqrt{\left(\sqrt{2}+1\right)^2}}\)

\(=\sqrt{13+3\cdot\left(\sqrt{2}+1\right)}\)

\(=\sqrt{13+3\sqrt{2}+3}\)

\(=\sqrt{16+3\sqrt{2}}\)

Ta có: \(C=\frac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)

\(=\frac{2\sqrt{3-\sqrt{3+\sqrt{12+2\cdot\sqrt{12}\cdot1+1}}}}{\sqrt{6}-\sqrt{2}}\)

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

\(=\frac{2\sqrt{3-\sqrt{3+\sqrt{12}+1}}}{\sqrt{6}-\sqrt{2}}\)

\(=\frac{2\sqrt{3-\sqrt{3+2\cdot\sqrt{3}\cdot1+1}}}{\sqrt{6}-\sqrt{2}}\)

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

\(=\frac{2\cdot\sqrt{3-\sqrt{3}-1}}{\sqrt{6}-\sqrt{2}}\)

\(=\frac{2\cdot\sqrt{2-\sqrt{3}}}{\sqrt{2}\cdot\left(\sqrt{3}-1\right)}=\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{3}-1}\)

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

Ta có: \(D=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{18-\sqrt{128}}}}}\)

\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{16-2\cdot4\cdot\sqrt{2}+2}}}}\)

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

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

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

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

\(=\sqrt{6+2\sqrt{2}\sqrt{1}}\)

\(=\sqrt{6+2\sqrt{2}}\)

Đề thiếu nha:

\(E=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)

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

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

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

\(=\frac{2\sqrt{3+\sqrt{3-2\sqrt{3}+1}}}{\sqrt{6}+\sqrt{2}}\)

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

\(=\frac{2\sqrt{3+\sqrt{3}-1}}{\sqrt{2}\left(\sqrt{3}+1\right)}\)(vì \(\sqrt{3}>1\))

\(=\frac{\sqrt{2}.\sqrt{2+\sqrt{3}}}{\sqrt{3}+1}\)

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

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

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

\(D=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)

\(\Rightarrow D\sqrt{2}=\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}\)

\(=\sqrt{5+2\sqrt{15}+3}+\sqrt{5-2\sqrt{15}+3}-2\sqrt{5-2\sqrt{5}+1}\)

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

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

\(=2\sqrt{5}-2\sqrt{5}+2=2\)

\(\Rightarrow D=\frac{2}{\sqrt{2}}=\sqrt{2}\)

\(H=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)

\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{16-8\sqrt{2}+2}}}}\)

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

\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}\)(vì \(4>\sqrt{2}\))

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

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

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

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

\(a.A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

⇔ \(A^2=\) \(\left(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\right)^2\)

⇔ \(A^2=4+\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}+4-\sqrt{10+2\sqrt{5}}\)⇔ \(A^2=8+2\sqrt{16-10-2\sqrt{5}}=8+2\sqrt{5-2\sqrt{5}+1}=8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)

⇔ \(A^2=8+2\text{|}\sqrt{5}-1\text{|}\)

⇔ \(A^2=6+2\sqrt{5}=5+2\sqrt{5}+1=\left(\sqrt{5}+1\right)^2\)

⇔ \(\text{ |}A\text{ |}=\text{ |}\sqrt{5}+1\text{ |}\)

⇔ \(A=\sqrt{5}+1\)

\(A=\frac{\left(\sqrt{3}+\sqrt{2}\right)^2\left(5-2\sqrt{6}\right)^2\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{9\sqrt{3}-11\sqrt{2}}=\frac{\left(\sqrt{3}+\sqrt{2}\right)^2\left(\sqrt{3}-\sqrt{2}\right)\left(5-2\sqrt{6}\right)^2}{9\sqrt{3}-11\sqrt{2}}\)

\(=\left(\sqrt{3}+\sqrt{2}\right)\left(9\sqrt{3}+11\sqrt{3}\right)\left(5-2\sqrt{6}\right)^2\)

\(=\left(49+20\sqrt{6}\right)\left(5-2\sqrt{6}\right)^2=\left(5+2\sqrt{6}\right)^2\left(5-2\sqrt{6}\right)^2=1\)

\(A=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)

\(=\sqrt{4+5}=3\)

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

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

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

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

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

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

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

\(A=\left(2-\sqrt{3}\right)\sqrt{4+2.2.\sqrt{3}+3}=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=1\)

các câu còn lại làm tương tự nhé bạn !

a: \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)

\(=4-\sqrt{15}+\sqrt{15}=4\)

b: \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)

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

\(=2\sqrt{3}\)

c: \(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)

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

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