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5 tháng 10 2020
b) \(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}=\sqrt{4+5}=3\)
PM
1
3 tháng 7 2017
\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
\(=\sqrt{6+2\sqrt{5-\sqrt{13+4\sqrt{3}}}}\)
\(=\sqrt{6+2\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{6+4\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
M
1
12 tháng 7 2017
\(\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{5-\sqrt{13+2\sqrt{12}}}=\sqrt{5-\sqrt{12+2\sqrt{12}\sqrt{1}+1}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}=\sqrt{5-\left(\sqrt{12}+1\right)}=\sqrt{5-\sqrt{12}-1}=\sqrt{3+2\sqrt{3}+1}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
NC
4 tháng 10 2020
a) \(\sqrt{\sqrt{5}-\sqrt{3}-\sqrt{29-6\sqrt{20}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3}-\sqrt{\left(\sqrt{20}-3\right)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3}-2\sqrt{5}+3}\)
\(=\sqrt{3-\sqrt{3}-\sqrt{5}}\)
L
25 tháng 11 2020
Cho sửa phần mẫu số của câu trên thành \(\sqrt{6}+\sqrt{2}\)
\(\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-|2\sqrt{3}+1|}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{4+2\sqrt{3}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+|\sqrt{3}-1|}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{2+\sqrt{3}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{\sqrt{2}.\sqrt{4+2\sqrt{3}}}{\sqrt{2}\left(\sqrt{3}+1\right)}\)
\(=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{3}+1}\)
\(=\frac{\sqrt{3}+1}{\sqrt{3}+1}=1\)
WR
23 tháng 6 2019
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
16 tháng 9 2018
(=)G 3= \(5+2\sqrt{13}+5-2\sqrt{13}+3\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}.G\)
(=) G3 = 10 + 3\(\sqrt[3]{25-52}.G\)= 10 -9G (=) G3 + 9G -10 =0 (=) (G-1)(G2 + G +10)= 0 => G=1 ( G2 + G +10\(\ne0\) )
24 tháng 4 2017
a) \(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
\(=2\sqrt{40.2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5.4\sqrt{3}}\)
\(=\left(2\sqrt{80}-2\sqrt{5}-3\sqrt{20}\right).\sqrt{\sqrt{3}}\)
\(=\left(8\sqrt{5}-2\sqrt{5}-6\sqrt{5}\right).\sqrt{\sqrt{3}}=0\)
b) \(2\sqrt{8\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
\(=\left(4\sqrt{2}-2\sqrt{5}-6\sqrt{5}\right).\sqrt{\sqrt{3}}\)
\(=\left(4\sqrt{2}-8\sqrt{5}\right).\sqrt{\sqrt{3}}\)
\(=\sqrt{\sqrt{3}}\left(\sqrt{2}-2\sqrt{5}\right)\)
29 tháng 8 2019
a/ \(\sqrt{2}+\sqrt{6}\)
b/ Sửa đề:
\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}=1\)
c/ \(1+\sqrt{2}+\sqrt{5}\)
25 tháng 6 2018
\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}.\)
\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{3+4+2\sqrt{12}}}}\)
\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(\sqrt{3}+\sqrt{4}\right)^2}}}\)
\(\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)
\(\sqrt{5\sqrt{3}+5\sqrt{25+3-2.\sqrt{25.3}}}\)
\(\sqrt{5\sqrt{3}+5\sqrt{\left(\sqrt{25}-\sqrt{3}\right)^2}}\)
\(\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)
\(\sqrt{25}=5\)
\(\sqrt{3+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
\(=\sqrt{3+2\sqrt{5-\sqrt{13+2\sqrt{12}}}}\)
\(=\sqrt{3+2\sqrt{5-\left(\sqrt{12}+1\right)}}\)
\(=\sqrt{3+2\sqrt{4-2\sqrt{3}}}=\sqrt{3+2\left(\sqrt{3}-1\right)}\)
\(=\sqrt{1+2\sqrt{3}}\)
\(\sqrt{3+2\sqrt{5-\sqrt{13+2\sqrt{12}}}}\)
=\(\sqrt{3+2\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}\)
=\(\sqrt{3+2\sqrt{5-2\sqrt{3}-1}}\)
=\(\sqrt{3+2\sqrt{4-2\sqrt{3}}}\)
=\(\sqrt{3+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
=\(\sqrt{3+2.\left(\sqrt{3}-1\right)}\)
=\(\sqrt{2\sqrt{3}+1}\)