\(\sqrt{13+4\sqrt{3}}\)\(+\sqrt{13-4\sqrt{3}}\)\(TINH\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
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, \(\sqrt{5-\sqrt{13+4\sqrt{3}}}+\sqrt{3+\sqrt{13+4\sqrt{3}}}\)
\(\Leftrightarrow\sqrt{5-\sqrt{\left(2\sqrt{3}\right)^2+2.2\sqrt{3}+1}}+\sqrt{3+\sqrt{\left(2\sqrt{3}\right)^2+2.2\sqrt{3}+1}}\)
\(\Leftrightarrow\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}+\sqrt{3+\sqrt{\left(2\sqrt{3}+1\right)^2}}\)
\(\Leftrightarrow\sqrt{5-2\sqrt{3}-1}+\sqrt{3+2\sqrt{3}+1}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(\Leftrightarrow\sqrt{3}-1+\sqrt{3}+1\)
\(\Leftrightarrow2\sqrt{3}\)
14 tháng 7 2019
\(\sqrt{24+8\sqrt{5}}+\) \(\sqrt{9-4\sqrt{5}}=\) \(\sqrt{\left(2\sqrt{5}\right)^2+2.2\sqrt{5}.2+4}\) + \(\sqrt{5-2\sqrt{5}.2+4}\)
= \(\sqrt{\left(2\sqrt{5}+2\right)^2}+\) \(\sqrt{\left(\sqrt{5}-2\right)^2}\) = \(2\sqrt{5}+2+\sqrt{5}-2=3\sqrt{5}\)
==================================================
\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\) = \(\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)= \(\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)
===========================================================
\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)
= \(\sqrt{13+30\sqrt{3+2\sqrt{2}}}=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{43+30\sqrt{2}}\) \(=\sqrt{\left(3\sqrt{2}+5\right)^2}=3\sqrt{2}+5\)
================================================================
AT
30 tháng 6 2018
a/ \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
\(=\sqrt{13+30\sqrt{2+\sqrt{1+2\cdot1\cdot2\sqrt{2}+8}}}\)
\(=\sqrt{13+30\sqrt{2+\sqrt{\left(1+2\sqrt{2}\right)^2}}}\)
\(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\)
\(=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{43+30\sqrt{2}}\)
\(=\sqrt{25+2\cdot5\cdot3\sqrt{2}+18}=\sqrt{\left(5+3\sqrt{2}\right)^2}=5+3\sqrt{2}\)
AT
30 tháng 6 2018
b/ \(\left(\sqrt{3}-\sqrt{2}\right)\cdot\sqrt{5+2\sqrt{6}}=\sqrt{3\left(5+2\sqrt{6}\right)}-\sqrt{2\left(5+2\sqrt{6}\right)}\)
\(=\sqrt{15+6\sqrt{6}}-\sqrt{10+4\sqrt{6}}\)
\(=\sqrt{\left(3+\sqrt{6}\right)^2}-\sqrt{\left(2+\sqrt{6}\right)^2}\)
\(=3+\sqrt{6}-2-\sqrt{6}=1\)
c/ \(\sqrt{5-\sqrt{13+4\sqrt{3}}}+\sqrt{3+\sqrt{13+4\sqrt{3}}}\)
\(=\sqrt{5-\sqrt{\left(1+2\sqrt{3}\right)^2}}+\sqrt{3+\sqrt{\left(1+2\sqrt{3}\right)^2}}\)
\(=\sqrt{5-1-2\sqrt{3}}+\sqrt{3+1+2\sqrt{3}}\)
\(=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(=\sqrt{3}-1+1+\sqrt{3}=2\sqrt{3}\)
7 tháng 8 2018
\(a,=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\left(\sqrt{20}-3\right)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-\sqrt{20}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}\)
\(=\sqrt{1}=1\)
b,c
\(\sqrt{13+4\sqrt{3}}=\sqrt{13+2\sqrt{12}}=\sqrt{12}+1=2\sqrt{3}+1\)
=>BT=\(\sqrt{5-\left(2\sqrt{3}+1\right)}+\sqrt{3+\left(2\sqrt{3}+1\right)}\)
\(=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)
c,\(=\sqrt{1+\sqrt{3+2\sqrt{3}+1}}+\sqrt{1-\sqrt{3-\left(2\sqrt{3}-1\right)}}\)
\(=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
MN
8 tháng 7 2020
Sửa đề nha :
Đặt
\(A=\sqrt{1+\sqrt{3+\sqrt{13+4\sqrt{3}}}}+\sqrt{1-\sqrt{3-\sqrt{13-4\sqrt{3}}}}\)
\(A=\sqrt{1+\sqrt{3+\sqrt{\left(2\sqrt{3}+1\right)^2}}}+\sqrt{1-\sqrt{3-\sqrt{\left(2\sqrt{3}-1\right)^2}}}\)
\(A=\sqrt{1+\sqrt{4+2\sqrt{3}}}+\sqrt{1-\sqrt{4-2\sqrt{3}}}\)
\(A=\sqrt{1+\sqrt{\left(\sqrt{3}+1\right)^2}}+\sqrt{1-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(A=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(A^2=2+\sqrt{3}+2-\sqrt{3}+2\sqrt{2+\sqrt{3}+2-\sqrt{3}}\)
\(A^2=4+2\sqrt{4}=6\)
\(A=\sqrt{6}\)
Vậy ....
\(\)
MN
8 tháng 7 2020
Sửa từ dòng 6 :
\(A^2=2+\sqrt{3}+2-\sqrt{3}+2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\)
\(A^2=4+2\sqrt{1}=6\)
\(A=6\)
Vậy ...
23 tháng 8 2020
Ta có: \(\sqrt{1+\sqrt{3+\sqrt{13+4\sqrt{3}}}}+\sqrt{1-\sqrt{3-\sqrt{13-4\sqrt{3}}}}\)
\(=\sqrt{1+\sqrt{3+\sqrt{12+2\cdot2\sqrt{3}\cdot1+1}}}+\sqrt{1-\sqrt{3-\sqrt{12-2\cdot2\sqrt{3}\cdot1+1}}}\)
\(=\sqrt{1+\sqrt{3+\sqrt{\left(2\sqrt{3}+1\right)^2}}}+\sqrt{1-\sqrt{3-\sqrt{\left(2\sqrt{3}-1\right)^2}}}\)
\(=\sqrt{1+\sqrt{3+\left|2\sqrt{3}+1\right|}}+\sqrt{1-\sqrt{3-\left|2\sqrt{3}-1\right|}}\)
\(=\sqrt{1+\sqrt{3+2\sqrt{3}+1}}+\sqrt{1-\sqrt{3-\left(2\sqrt{3}-1\right)}}\)(Vì \(2\sqrt{3}>1>0\))
\(=\sqrt{1+\sqrt{4+2\sqrt{3}}}+\sqrt{1-\sqrt{3-2\sqrt{3}+1}}\)
\(=\sqrt{1+\sqrt{3+2\cdot\sqrt{3}\cdot1+1}}+\sqrt{1-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{1+\sqrt{\left(\sqrt{3}+1\right)^2}}+\sqrt{1-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{1+\left|\sqrt{3}+1\right|}+\sqrt{1-\left|\sqrt{3}-1\right|}\)
\(=\sqrt{1+\sqrt{3}+1}+\sqrt{1-\left(\sqrt{3}-1\right)}\)(Vì \(\sqrt{3}>1>0\))
\(=\sqrt{2+\sqrt{3}}+\sqrt{1-\sqrt{3}+1}\)
\(=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(=\frac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)
\(=\frac{\sqrt{3+2\cdot\sqrt{3}\cdot1+1}+\sqrt{3-2\cdot\sqrt{3}\cdot1+1}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}\)
\(=\frac{\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|}{\sqrt{2}}\)
\(=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}\)(Vì \(\sqrt{3}>1>0\))
\(=\frac{2\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{12}}{\sqrt{2}}=\sqrt{6}\)
24 tháng 8 2023
a: \(=\sqrt{8+2\cdot2\sqrt{2}\cdot\sqrt{5}+5}+\sqrt{8-2\cdot2\sqrt{2}\cdot\sqrt{5}+5}\)
\(=\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
\(=2\sqrt{2}+\sqrt{5}+2\sqrt{2}-\sqrt{5}=4\sqrt{2}\)
b: \(=2\cdot\sqrt{17-3\sqrt{32}}\)
\(=2\cdot\sqrt{9-2\cdot3\cdot2\sqrt{2}+8}\)
\(=2\left(3-2\sqrt{2}\right)=6-4\sqrt{2}\)
HC
2
TH
16 tháng 7 2018
\(\sqrt{5+2\sqrt{6}}-\sqrt{13-4\sqrt{3}}=\sqrt{3}+\sqrt{2}-\left(2\sqrt{3}-1\right)=1+\sqrt{2}-\sqrt{3}\)
9 tháng 7 2020
c, Ta có : \(\sqrt{13+4\sqrt{3}}=\sqrt{12+2\sqrt{12}+1}=\sqrt{\left(\sqrt{12}+1\right)^2}=\sqrt{12}+1\)
=> \(\sqrt{5-\sqrt{12}-1}+\sqrt{3+\sqrt{12}+1}\)
\(=\sqrt{4-\sqrt{12}}+\sqrt{4+\sqrt{12}}=\sqrt{3-2\sqrt{3}+1}+\sqrt{3+2\sqrt{3}+1}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)\(=\sqrt{3}+1+\sqrt{3}-1=2\sqrt{3}\)
\(A^2=\left(\sqrt{13+4\sqrt{3}}+\sqrt{13-4\sqrt{3}}\right)^2\)
\(=13+4\sqrt{3}+13-4\sqrt{3}+2\sqrt{\sqrt{13+4\sqrt{3}}\cdot\sqrt{13-4\sqrt{3}}}\)
\(=26+2\sqrt{13^2-\left(4\sqrt{3}\right)^2}\)
\(=26+2\sqrt{121}=26+22=48\)
\(\Rightarrow A^2=48\Rightarrow A=\sqrt{48}\)