\(\sqrt{\sqrt{5-\sqrt{3-\sqrt{29-2\cdot3\cdot2\sqrt{5}}}}}=\)\(\sqrt{\sqrt{5-\sqrt{3-2\sqrt{5}+3}}}=\sqrt{\sqrt{5-\sqrt{6-2\sqrt{5}}}}\)=\(\sqrt{\sqrt{5-\sqrt{5}+1}}=\sqrt{1}=1\)
\(\sqrt{\sqrt{5-\sqrt{3-\sqrt{29-2\cdot3\cdot2\sqrt{5}}}}}=\)\(\sqrt{\sqrt{5-\sqrt{3-2\sqrt{5}+3}}}=\sqrt{\sqrt{5-\sqrt{6-2\sqrt{5}}}}\)=\(\sqrt{\sqrt{5-\sqrt{5}+1}}=\sqrt{1}=1\)
Tính:
a) A=\(2\sqrt{3+\sqrt{5-13+\sqrt{48}}}\)
b) B=\(\sqrt{\sqrt{15}-\sqrt{3-\sqrt{29-2\sqrt{180}}}}\)
\(\sqrt{29-2\sqrt{180}}-\sqrt{9+4\sqrt{5}}\)
đề bài rút gọn ak
mình đang cần gấp
\(\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}=\sqrt{5}-\sqrt{3\sqrt{\left(\sqrt{20-3}\right)^2}}\)
\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}\)
\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
a) A=\(\sqrt{\left(4-\sqrt{15}\right)^2+\sqrt{15}}\)
b) B=\(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1-\sqrt{3}\right)^2}\)
c) C=\(\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)
d)D=\(\sqrt{29+12\sqrt{5}-\sqrt{29-12\sqrt{5}}}\)
Rút Gọn :
\(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)
\(C=\sqrt[3]{4^3+29\sqrt{2}}+\sqrt[3]{4^5-29\sqrt{2}}\)
a)\(\sqrt{29-12\sqrt{5}}\)
b) \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
1)Tính:
a) \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
b) \(\sqrt{6+2\sqrt{5}-\sqrt{29-12\sqrt{5}}}\)
c)\(\sqrt{2 +\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\times\sqrt{2004-2\sqrt{2006}-2\sqrt{2005}}\)
sắp xếp theo thứ tự tăng dần
a.3\(\sqrt{5}\);2\(\sqrt{6}\);\(\sqrt{29}\)và 4\(\sqrt{2}\)
b.5\(\sqrt{2}\);\(\sqrt{39}\);3\(\sqrt{8}\)và \(2\sqrt{15}\)