NA
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
23 tháng 10 2016
c/ = \(\sqrt{13+30\sqrt{2+\sqrt{8+2.2\sqrt{2}+1}}}\)
\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}\)
\(=\sqrt{43+30\sqrt{2}}\)
\(=\sqrt{25+2.3.5.\sqrt{2}+18}\)
\(=5+3\sqrt{2}\)
23 tháng 10 2016
d/ \(=\sqrt{13+6\sqrt{4+\sqrt{9-4\sqrt{2}}}}\)
\(=\sqrt{13+6\sqrt{4+2\sqrt{2}-1}}\)
\(=\sqrt{13+6\left(\sqrt{3}+1\right)}\)
\(=\sqrt{19+6\sqrt{2}}\)
\(=3\sqrt{2}+1\)
KT
6 tháng 7 2018
\(\sqrt{10+2\sqrt{17-4\sqrt{9+4\sqrt{5}}}}\)
\(=\sqrt{10+2\sqrt{17-4\sqrt{\left(\sqrt{5}+2\right)^2}}}\)
\(=\sqrt{10+2\sqrt{17-4\left(\sqrt{5}+2\right)}}\)
\(=\sqrt{10+2\sqrt{9-4\sqrt{5}}}\)
\(=\sqrt{10+2\sqrt{\left(\sqrt{5}-2\right)^2}}\)
\(=\sqrt{10+2\left(\sqrt{5}-2\right)}\)
\(=\sqrt{6+2\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=\sqrt{5}+1\)
KT
6 tháng 7 2018
\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
\(=\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\)
\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}\)
\(=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\)
\(=\sqrt{13+30\left(\sqrt{2}+1\right)}\)
\(=\sqrt{43+30\sqrt{2}}\)
\(=\sqrt{\left(3\sqrt{2}+5\right)^2}=3\sqrt{2}+5\)
1 tháng 7 2017
a,\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}\\ =\sqrt{13+30\left(\sqrt{2}+1\right)}\)
\(=\sqrt{43+30\sqrt{2}}=5+3\sqrt{2}\)
1 tháng 7 2017
b, \(\sqrt{5-\sqrt{13+4\sqrt{3}}}+\sqrt{3+\sqrt{13+4\sqrt{3}}}\)
\(\Leftrightarrow\sqrt{5-\sqrt{\left(2\sqrt{3}\right)^2+2.2\sqrt{3}+1}}+\sqrt{3+\sqrt{\left(2\sqrt{3}\right)^2+2.2\sqrt{3}+1}}\)
\(\Leftrightarrow\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}+\sqrt{3+\sqrt{\left(2\sqrt{3}+1\right)^2}}\)
\(\Leftrightarrow\sqrt{5-2\sqrt{3}-1}+\sqrt{3+2\sqrt{3}+1}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(\Leftrightarrow\sqrt{3}-1+\sqrt{3}+1\)
\(\Leftrightarrow2\sqrt{3}\)
6 tháng 6 2018
a/ \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(3-2\sqrt{5}\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=1\)
b,c tương tự
14 tháng 7 2019
\(\sqrt{24+8\sqrt{5}}+\) \(\sqrt{9-4\sqrt{5}}=\) \(\sqrt{\left(2\sqrt{5}\right)^2+2.2\sqrt{5}.2+4}\) + \(\sqrt{5-2\sqrt{5}.2+4}\)
= \(\sqrt{\left(2\sqrt{5}+2\right)^2}+\) \(\sqrt{\left(\sqrt{5}-2\right)^2}\) = \(2\sqrt{5}+2+\sqrt{5}-2=3\sqrt{5}\)
==================================================
\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\) = \(\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)= \(\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)
===========================================================
\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)
= \(\sqrt{13+30\sqrt{3+2\sqrt{2}}}=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{43+30\sqrt{2}}\) \(=\sqrt{\left(3\sqrt{2}+5\right)^2}=3\sqrt{2}+5\)
================================================================
28 tháng 7 2016
a) \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{20-2\cdot3\cdot\sqrt{20}+9}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{20}+3}}\)
\(=\sqrt{5-\sqrt{6-\sqrt{20}}}\)
\(=\sqrt{5-\sqrt{5-2\sqrt{5}+1}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{5}+1\right)^2}}\)
\(=\sqrt{5-\sqrt{5}-1}\)
\(=\sqrt{4-\sqrt{5}}\)
c)\(\left(\sqrt{3}-\sqrt{2}\right)\sqrt{5+2\sqrt{6}}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{3+2\cdot\sqrt{3}\cdot\sqrt{2}+2}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)\)
\(=3-2=1\)
d)\(\sqrt{5-\sqrt{13+4\sqrt{3}}}+\sqrt{3+\sqrt{13+4\sqrt{3}}}\)
\(=\sqrt{5-\sqrt{12+2\cdot\sqrt{12}+1}}+\sqrt{3+\sqrt{12+2\cdot\sqrt{12}+1}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}+\sqrt{3+\sqrt{\left(\sqrt{12}+1\right)^2}}\)
\(=\sqrt{5-\sqrt{12}-1}+\sqrt{3+\sqrt{12}+1}\)
\(=\sqrt{4-\sqrt{12}}+\sqrt{4+\sqrt{12}}\)
\(=\sqrt{3-2\sqrt{3}+1}+\sqrt{4+2\sqrt{3}+1}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\sqrt{3}-1+\sqrt{3+1}\)
\(=2\sqrt{3}\)
NV
27 tháng 7 2015
1. \(=\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}+\sqrt{3+\sqrt{\left(2\sqrt{3}+1\right)^2}}=\sqrt{5-2\sqrt{3}-1}+\sqrt{3+2\sqrt{3}+1}=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)
DK
21 tháng 6 2016
1/ \(\sqrt{5-\sqrt{13+4\sqrt{3}}}+\sqrt{3+\sqrt{13+4\sqrt{3}}}\)
\(=\sqrt{5-\left(1+\sqrt{12}\right)^2}+\sqrt{3+\left(1+\sqrt{12}\right)^2}\)
\(=\sqrt{5-\left|1+\sqrt{12}\right|}+\sqrt{3+\left|1+\sqrt{12}\right|}\)
\(=\sqrt{5-1-\sqrt{12}}+\sqrt{3+1+\sqrt{12}}\)
\(=\sqrt{4-\sqrt{12}}+\sqrt{4+\sqrt{12}}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left|\sqrt{3}-1\right|+\left|\sqrt{3}+1\right|\)
\(=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)
\(\sqrt{13+6\sqrt{4+\sqrt{9-4\sqrt{2}}}}\)
\(=\sqrt{13+6\sqrt{4+\sqrt{\left(2\sqrt{2}-1\right)^2}}}\)
\(=\sqrt{13+6\sqrt{4+\sqrt{\left|2\sqrt{2}-1\right|}}}\)
\(=\sqrt{13+6\sqrt{4+2\sqrt{2}-1}}\)(do \(2\sqrt{2}>1\))
\(=\sqrt{13+6\sqrt{3+2\sqrt{2}}}\)
\(=\sqrt{13+6\sqrt{\left(\sqrt{2}+1\right)^2}}\)
\(=\sqrt{13+6\left|\sqrt{2}+1\right|}\)
\(=\sqrt{13+6\left(\sqrt{2}+1\right)}\)(do \(\sqrt{2};1>0\))
\(=\sqrt{13+6\sqrt{2}+6}\)
\(=\sqrt{19+6\sqrt{2}}\)
\(=\sqrt{\left(3\sqrt{2}+1\right)^2}\)
\(=\left|3\sqrt{2}+1\right|\)
\(=3\sqrt{2}+1\)(do \(3\sqrt{2};1>0\))