Question

[Trần Thị Hảo]

Bài 1:

a) \(\sqrt{13-2\sqrt{42}}\)

b) \(\sqrt{46+6\sqrt{5}}\)

c) \(\sqrt{12-3\sqrt{15}}\)

d) \(\sqrt{11+\sqrt{96}}\)

Bài 2:

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

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

c) \(C=\sqrt{3-\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)\left(3+\sqrt{5}\right)\)

d) \(D=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)

e) \(E=\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)

g) \(G=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)

h) \(H=4x-\sqrt{9x^2-12x+4}\)

i) \(\frac{\sqrt{7}-\sqrt{2}}{\sqrt{7}+\sqrt{2}}+\frac{\sqrt{7}+\sqrt{2}}{\sqrt{7}-\sqrt{2}}\)

Answer 1

[Akai Haruma]

Bài 1:

a)

\(\sqrt{13-2\sqrt{42}}=\sqrt{6+7-2\sqrt{6.7}}=\sqrt{(\sqrt{7}-\sqrt{6})^2}=|\sqrt{7}-\sqrt{6}|=\sqrt{7}-\sqrt{6}\)

b)

\(\sqrt{46+6\sqrt{5}}=\sqrt{46+2\sqrt{45}}=\sqrt{45+1+2\sqrt{45.1}}=\sqrt{(\sqrt{45}+1)^2}=\sqrt{45}+1\)

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

c)

\(\sqrt{12-3\sqrt{15}}=\sqrt{\frac{24-6\sqrt{15}}{2}}=\sqrt{\frac{24-2\sqrt{135}}{2}}=\sqrt{\frac{15+9-2\sqrt{15.9}}{2}}\)

\(=\sqrt{\frac{(\sqrt{15}-\sqrt{9})^2}{2}}=\frac{\sqrt{15}-\sqrt{9}}{\sqrt{2}}=\frac{\sqrt{15}-3}{\sqrt{2}}\)

d)

\(\sqrt{11+\sqrt{96}}=\sqrt{11+2\sqrt{24}}=\sqrt{8+3+2\sqrt{8.3}}\)

\(=\sqrt{(\sqrt{8}+\sqrt{3})^2}=\sqrt{8}+\sqrt{3}\)

Answer 2

[Akai Haruma]

Bài 1:

a)

\(\sqrt{13-2\sqrt{42}}=\sqrt{6+7-2\sqrt{6.7}}=\sqrt{(\sqrt{7}-\sqrt{6})^2}=|\sqrt{7}-\sqrt{6}|=\sqrt{7}-\sqrt{6}\)

b)

\(\sqrt{46+6\sqrt{5}}=\sqrt{46+2\sqrt{45}}=\sqrt{45+1+2\sqrt{45.1}}=\sqrt{(\sqrt{45}+1)^2}=\sqrt{45}+1\)

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

c)

\(\sqrt{12-3\sqrt{15}}=\sqrt{\frac{24-6\sqrt{15}}{2}}=\sqrt{\frac{24-2\sqrt{135}}{2}}=\sqrt{\frac{15+9-2\sqrt{15.9}}{2}}\)

\(=\sqrt{\frac{(\sqrt{15}-\sqrt{9})^2}{2}}=\frac{\sqrt{15}-\sqrt{9}}{\sqrt{2}}=\frac{\sqrt{15}-3}{\sqrt{2}}\)

d)

\(\sqrt{11+\sqrt{96}}=\sqrt{11+2\sqrt{24}}=\sqrt{8+3+2\sqrt{8.3}}\)

\(=\sqrt{(\sqrt{8}+\sqrt{3})^2}=\sqrt{8}+\sqrt{3}\)