Question

[EDOGAWA CONAN]

Tính thu gọn :

a , \(\sqrt{17-12\sqrt{2}}-\sqrt{17+12\sqrt{2}}\)

b , \(\sqrt{27+12\sqrt{5}}-\sqrt{27-12\sqrt{5}}\)

c , \(\sqrt{15-6\sqrt{6}}+\sqrt{15+\sqrt{6\sqrt{6}}}\)

d , \(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)

e , \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)

f , \(\sqrt{5+\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

Answer 1

[Nguyễn Tấn An]

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

Answer 2

[EDOGAWA CONAN]

mik sửa lại câu f , tí nhé :

f , \(\sqrt{\sqrt{5}+\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)