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TH
1
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)\(\sqrt{5-\sqrt{13+4\sqrt{3}}}+\sqrt{3+\sqrt{13+4\sqrt{3}}}\)
\(=\sqrt{5-\sqrt{12+2\cdot\sqrt{12}+1}}+\sqrt{3+\sqrt{12+2\cdot\sqrt{12}+1}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}+\sqrt{3+\sqrt{\left(\sqrt{12}+1\right)^2}}\)
\(=\sqrt{5-\sqrt{12}-1}+\sqrt{3+\sqrt{12}+1}\)
\(=\sqrt{4-\sqrt{12}}+\sqrt{4+\sqrt{12}}\)
\(=\sqrt{3-2\sqrt{3}+1}+\sqrt{4+2\sqrt{3}+1}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\sqrt{3}-1+\sqrt{3+1}\)
\(=2\sqrt{3}\)
14 tháng 7 2019
\(\sqrt{24+8\sqrt{5}}+\) \(\sqrt{9-4\sqrt{5}}=\) \(\sqrt{\left(2\sqrt{5}\right)^2+2.2\sqrt{5}.2+4}\) + \(\sqrt{5-2\sqrt{5}.2+4}\)
= \(\sqrt{\left(2\sqrt{5}+2\right)^2}+\) \(\sqrt{\left(\sqrt{5}-2\right)^2}\) = \(2\sqrt{5}+2+\sqrt{5}-2=3\sqrt{5}\)
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\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\) = \(\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)= \(\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)
===========================================================
\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)
= \(\sqrt{13+30\sqrt{3+2\sqrt{2}}}=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{43+30\sqrt{2}}\) \(=\sqrt{\left(3\sqrt{2}+5\right)^2}=3\sqrt{2}+5\)
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B
1
24 tháng 10 2021
\(=\dfrac{\sqrt{5}-2}{5-4}-\sqrt{\left(\sqrt{5}+2\right)^2}=\sqrt{5}-2-\sqrt{5}-2=-4\)
NT
1
27 tháng 7 2021
a) \(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)
\(=2\sqrt{5}+2+\sqrt{5}-2\)
\(=3\sqrt{5}\)
b) \(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\)
\(=3-2\sqrt{2}+2\sqrt{2}-1\)
=2
c) \(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)
\(=2-\sqrt{2}+3\sqrt{2}-2\)
\(=2\sqrt{2}\)
AH
Akai Haruma
Giáo viên
26 tháng 8 2023
Lời giải:
a.
\(=\sqrt{5+2.2\sqrt{5}+2^2}-\sqrt{5-2.2\sqrt{5}+2^2}\)
$=\sqrt{(\sqrt{5}+2)^2}-\sqrt{(\sqrt{5}-2)^2}$
$=|\sqrt{5}+2|-|\sqrt{5}-2|=(\sqrt{5}+2)-(\sqrt{5}-2)=4$
b.
$=\sqrt{3-2.3\sqrt{3}+3^2}+\sqrt{3+2.3.\sqrt{3}+3^2}$
$=\sqrt{(\sqrt{3}-3)^2}+\sqrt{(\sqrt{3}+3)^2}$
$=|\sqrt{3}-3|+|\sqrt{3}+3|$
$=(3-\sqrt{3})+(\sqrt{3}+3)=6$
c.
$=\sqrt{2+2.3\sqrt{2}+3^2}-\sqrt{2-2.3\sqrt{2}+3^2}$
$=\sqrt{(\sqrt{2}+3)^2}-\sqrt{(\sqrt{2}-3)^2}$
$=|\sqrt{2}+3|-|\sqrt{2}-3|$
$=(\sqrt{2}+3)-(3-\sqrt{2})=2\sqrt{2}$
29 tháng 6 2018
\(\sqrt{4-\sqrt{9+4\sqrt{2}}}=\sqrt{4-\sqrt{1+2.2.\sqrt{2}+\left(2\sqrt{2}\right)^2}}\)
\(=\sqrt{4-\sqrt{\left(1+2\sqrt{2}\right)^2}}\)
\(=\sqrt{4-\left|1+2\sqrt{2}\right|}=\sqrt{4-1-2\sqrt{2}}=\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{1-2.\sqrt{2}.1+\left(\sqrt{2}\right)^2}=\sqrt{\left(1-\sqrt{2}\right)^2}=\left|1-\sqrt{2}\right|=\sqrt{2}-1\)
Vậy ....
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, \(\sqrt{5-\sqrt{13+4\sqrt{3}}}+\sqrt{3+\sqrt{13+4\sqrt{3}}}\)
\(\Leftrightarrow\sqrt{5-\sqrt{\left(2\sqrt{3}\right)^2+2.2\sqrt{3}+1}}+\sqrt{3+\sqrt{\left(2\sqrt{3}\right)^2+2.2\sqrt{3}+1}}\)
\(\Leftrightarrow\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}+\sqrt{3+\sqrt{\left(2\sqrt{3}+1\right)^2}}\)
\(\Leftrightarrow\sqrt{5-2\sqrt{3}-1}+\sqrt{3+2\sqrt{3}+1}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(\Leftrightarrow\sqrt{3}-1+\sqrt{3}+1\)
\(\Leftrightarrow2\sqrt{3}\)
NH
21 tháng 5 2023
a)
\(M=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{4+4\sqrt{5}+5}-\sqrt{4-4\sqrt{5}+5}\)
\(=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=\left|2+\sqrt{5}\right|-\left|2-\sqrt{5}\right|\)
\(=2+\sqrt{5}-\left(\sqrt{5}-2\right)\) (vì \(2+2\sqrt{5}>0;2-\sqrt{5}< 0\) )
\(=2+\sqrt{5}-\sqrt{5}+2\\ =4\)
b)
\(N=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\left|\sqrt{7}-1\right|-\left|\sqrt{7}+1\right|\)
\(=\sqrt{7}-1-\left(\sqrt{7}+1\right)\) (vì \(\sqrt{7}-1>0;\sqrt{7}+1>0\) )
\(=\sqrt{7}-1-\sqrt{7}-1\\ =-2\)
\(13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}\)
\(=13+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}\)
\(=13+30\sqrt{3+2\sqrt{2}}=13+30\sqrt{\left(\sqrt{2}+1\right)^2}\)
\(=13+30\sqrt{2}+30=43+30\sqrt{2}=43+2.5.3\sqrt{2}\)
\(=25+2.5.3\sqrt{2}+18=\left(5+2\sqrt{2}\right)^2\)