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\(\sqrt{\sqrt{\left(3\right)^8}}\)=\(\sqrt{\sqrt{6561}}=\sqrt{81}=9\)
\(\sqrt[2]{\left(-5^4\right)}=\sqrt[2]{625}=25\)
1.
\(\sqrt{\frac{2+\sqrt{3}}{2}}\\ =\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2}}\\ =\frac{\sqrt{4+2\sqrt{3}}}{2}\\ =\frac{\sqrt{\left(1+\sqrt{3}\right)^2}}{2}\\ =\frac{1+\sqrt{3}}{2}\)
2.
\(\sqrt{\frac{14+5\sqrt{3}}{2}}\\ =\frac{\sqrt{14+5\sqrt{3}}}{\sqrt{2}}\\ =\frac{\sqrt{28+10\sqrt{3}}}{2}\\ =\frac{\sqrt{\left(5+\sqrt{3}\right)^2}}{2}\\ =\frac{5+\sqrt{3}}{2}\)
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) \(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)
\(=2-\sqrt{2}+3\sqrt{2}-2\)
\(=2\sqrt{2}\)
\(\sqrt{10-4\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{2^2-2.2.\sqrt{6}+\left(\sqrt{6}\right)^2}+\sqrt{3^2-2.3.2\sqrt{6}+\left(2\sqrt{6}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{6}\right)^2}+\sqrt{\left(3-2\sqrt{6}\right)^2}\)
\(=-\left(2-\sqrt{6}\right)-\left(3-2\sqrt{6}\right)\)
\(=-2+\sqrt{6}-3+2\sqrt{6}\)
\(=-5+3\sqrt{6}\)
\(\sqrt{16-6\sqrt{7}}+\sqrt{32-8\sqrt{7}}\)
\(=\sqrt{3^2-2.3.\sqrt{7}+\left(\sqrt{7}\right)^2}+\sqrt{2^2-2.2.2\sqrt{7}+\left(2\sqrt{7}\right)^2}\)
\(=\sqrt{\left(3-\sqrt{7}\right)^2}+\sqrt{\left(2-2\sqrt{7}\right)^2}\)
\(=3-\sqrt{7}-\left(2-2\sqrt{7}\right)\)
\(=3-\sqrt{7}-2+2\sqrt{7}\)
\(=1+\sqrt{7}\)
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.
$=\sqrt{2+2.3\sqrt{2}+3^2}-\sqrt{2-2.3\sqrt{2}+3^2}$
$=\sqrt{(\sqrt{2}+3)^2}-\sqrt{(\sqrt{2}-3)^2}$
$=|\sqrt{2}+3|-|\sqrt{2}-3|$
$=(\sqrt{2}+3)-(3-\sqrt{2})=2\sqrt{2}$
\(1,\\ a,=\sqrt{\left(3+\sqrt{7}\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}=3+\sqrt{7}-\sqrt{7}+1=4\\ b,K=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\ c,=\sqrt{\left(6-2\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-4\right)^2}=6-2\sqrt{6}+2\sqrt{6}-4=2\\ e,=\sqrt{\left(2-\sqrt{2}\right)^2}-\left(\sqrt{6}-\sqrt{2}\right)=2-\sqrt{2}-\sqrt{6}+\sqrt{2}=2-\sqrt{6}\)
\(2,\\ a,A=\dfrac{x-3\sqrt{x}+3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{x+9}\\ A=\dfrac{x+9}{\left(\sqrt{x}-3\right)\left(x+9\right)}=\dfrac{1}{\sqrt{x}-3}\\ b,x=4+2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{3}+1\\ \Leftrightarrow A=\dfrac{1}{\sqrt{3}+1-3}=\dfrac{1}{\sqrt{3}+2}=2-\sqrt{3}\)
a ) Để ý thấy \(16\sqrt{3}=2.2\sqrt{3}.4=2.\sqrt{12}.4\) , như vậy , ta sẽ tách :
\(28=12+16\) \(\Rightarrow\sqrt{\sqrt{28+16\sqrt{3}}=\sqrt{\sqrt{12+16+16\sqrt{3}}}}=\sqrt{\sqrt{\left(\sqrt{12}+4\right)^2}}=\sqrt{\sqrt{12}+4}\)
\(=\sqrt{3+2.\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
b ) \(4\sqrt{3}=2.2\sqrt{3}\), tách \(7=4+3\)
c ) \(24\sqrt{5}=2.\sqrt{5}.12=2.\sqrt{5}.2.6=2.\sqrt{20}.6\) , tách : \(56=20+36\)
d ) \(2\sqrt{11}=2.11.1\) , tách : \(12=11+1\)
e ) \(4\sqrt{2}=2.\sqrt{2}.2.1=2.\sqrt{8}.1\) , tách : \(9=8+1\)
a) \(\sqrt{\sqrt{28+16\sqrt{3}}}\)
\(=\sqrt{\sqrt{\left(2\sqrt{3}\right)^2+2\cdot2\sqrt{3}\cdot4+16}}\)
\(=\sqrt{\sqrt{\left(2\sqrt{3}+4\right)^2}}\) \(=\sqrt{2\sqrt{3}+4}\)
\(=\sqrt{3+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
b)\(\sqrt{7+4\sqrt{3}}=\sqrt{4+4\sqrt{3}+3}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}=2+\sqrt{3}\)
c) \(\sqrt{\sqrt{56-24\sqrt{5}}}=\sqrt{\sqrt{\left(2\sqrt{5}\right)^2-2\cdot2\sqrt{5}\cdot6}+36}\)
\(=\sqrt{\sqrt{\left(2\sqrt{5}-6\right)^2}}=\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5-2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)
d) \(\sqrt{12-2\sqrt{11}}=\sqrt{11-2\sqrt{11}+1}\)
\(=\sqrt{\left(\sqrt{11}-1\right)^2}=\sqrt{11}-1\)
e) \(\sqrt{9+4\sqrt{2}}=\sqrt{\left(2\sqrt{2}\right)^2+2\cdot2\sqrt{2}+1}\)
\(=\sqrt{\left(2\sqrt{2}+1\right)^2}=2\sqrt{2}+1\)