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

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

\(a,\sqrt{29-12\sqrt{5}}=2\sqrt{5}-3\\ b,\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\\ =\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\\ =\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\\ =\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}\\ =\sqrt{1}=1\)

a: \(\sqrt{29-12\sqrt{5}}=2\sqrt{5}-3\)

b: \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

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

\(=\sqrt{\sqrt{5}-\sqrt{5}+1}\)

=1

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

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

b: \(A=2-\sqrt{3}+\sqrt{3}-1=1\)

c: \(C=3\sqrt{5}-2-3\sqrt{5}-2=-4\)

d: Sửa đề: \(D=\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)

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

=6

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

\(A=\left|4-\sqrt{15}\right|+\sqrt{15}\)

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

\(A=4\)

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

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

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

\(B=1\)

c) \(C=\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)

\(C=\sqrt{\left(3\sqrt{5}\right)^2-2\cdot3\sqrt{15}\cdot2+2^2}-\sqrt{\left(3\sqrt{5}\right)^2+2\cdot3\sqrt{5}\cdot2+2^2}\)

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

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

\(C=3\sqrt{5}-2-3\sqrt{5}-2\)

\(C=-4\)

d) \(D=\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)

\(D=\sqrt{\left(2\sqrt{5}\right)^2+2\cdot2\sqrt{5}\cdot3+3^2}-\sqrt{\left(2\sqrt{5}\right)^2-2\cdot2\sqrt{5}\cdot3+3^3}\)

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

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

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

\(D=6\)

a) Ta có: \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2\cdot\sqrt{20}\cdot3+9}}}\)

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

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

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

\(=\sqrt{\sqrt{5}-\sqrt{5-2\cdot\sqrt{5}\cdot1+1}}\)

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

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

\(=\sqrt{\sqrt{5}-\sqrt{5}+1}\)

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

b) Ta có: \(\sqrt{6+2\sqrt{5}-\sqrt{29-12\sqrt{5}}}\)

\(=\sqrt{6+2\sqrt{5}-\sqrt{20-2\cdot2\sqrt{5}\cdot3+9}}\)

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

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

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

c) Sửa đề: \(\sqrt{2+\sqrt{5+\sqrt{13-\sqrt{48}}}}\)

Ta có: \(\sqrt{2+\sqrt{5+\sqrt{13-\sqrt{48}}}}\)

\(=\sqrt{2+\sqrt{5+\sqrt{12-2\cdot2\sqrt{3}\cdot1+1}}}\)

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

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

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

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

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

d) Ta có: \(\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)

\(=\dfrac{\left(6-2\sqrt{5}\right)\sqrt{6+2\sqrt{5}}+\left(6+2\sqrt{5}\right)\sqrt{6-2\sqrt{5}}}{2\sqrt{2}}\)

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

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

\(=\dfrac{4\cdot2\sqrt{5}}{2\sqrt{2}}\)

\(=\dfrac{8\sqrt{5}}{2\sqrt{2}}=\dfrac{4\sqrt{5}}{\sqrt{2}}=2\sqrt{10}\)

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

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

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

Bài 1 : \(\sqrt{49-12\sqrt{5}}+\sqrt{49+12\sqrt{5}}\)

\(=\sqrt{45-4\sqrt{45}+4}+\sqrt{45+4\sqrt{45}+4}\)

\(=\sqrt{\left(\sqrt{45}-2\right)^2}+\sqrt{\left(\sqrt{45}+2\right)^2}\)

\(=\sqrt{45}-2+\sqrt{45}+2=2\sqrt{45}\)

Bài 2 : \(\sqrt{29+12\sqrt{5}}+\sqrt{29-12\sqrt{5}}\)

\(=\sqrt{20+6\sqrt{20}+9}+\sqrt{20-6\sqrt{20}+9}\)

\(=\sqrt{\left(\sqrt{20}+3\right)^2}+\sqrt{\left(\sqrt{20}-3\right)^2}\)

\(=\sqrt{20}+3+\sqrt{20}-3=2\sqrt{20}\)

Bài 3 : \(\sqrt{31-12\sqrt{3}}+\sqrt{31+12\sqrt{3}}\)

\(=\sqrt{27-4\sqrt{27}+4}+\sqrt{27+4\sqrt{27}+4}\)

\(=\sqrt{\left(\sqrt{27}-2\right)^2}+\sqrt{\left(\sqrt{27}+2\right)^2}\)

\(=\sqrt{27}-2+\sqrt{27}+2=2\sqrt{27}\)

Chúc bạn học tốt

4 , Ta có :

\(\sqrt{39-12\sqrt{3}}-\sqrt{39+12\sqrt{3}}\)

\(=\sqrt{3-2.6.\sqrt{3}+6^2}-\sqrt{3+2.6.\sqrt{3}+6^2}\)

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

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

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

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

1,\(\sqrt{4-2.2.3\sqrt{5}+45}+\sqrt{4+2.2.3\sqrt{5}+45}\)

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

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

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

\(=6\sqrt{5}\)

x = \(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)

x = \(\sqrt{\left(2\sqrt{5}\right)^2+2.2\sqrt{5}.3+3^2}\) - \(\sqrt{\left(2\sqrt{5}\right)^2-2.2\sqrt{5}.3+3^2}\)

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

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

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

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

\(x=\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)

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

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

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

\(\Rightarrow x=6\)

\(x=\sqrt{9+2.3.2\sqrt{5}+\left(2\sqrt{5}\right)^2}+\sqrt{9-2.3.2\sqrt{5}+\left(2\sqrt{5}\right)^2}\)

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

=> x = 6

\(=-10\)

\(=-6\)