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) \(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)

\(=2-\sqrt{2}+3\sqrt{2}-2\)

\(=2\sqrt{2}\)

a,\(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}=\sqrt{2^2+2\cdot2\cdot\left(2\sqrt{5}\right)+\left(2\sqrt{5}\right)^2}\) \(+\sqrt{\left(\sqrt{5}\right)^2-2\cdot2\sqrt{5}+2^2}=\sqrt{\left(2+2\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\)=\(2+2\sqrt{5}+\sqrt{5}-2=3\sqrt{5}\) 

b,\(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}=3-2\sqrt{2}+2\sqrt{2}+1=4\)

c,\(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-2\right)^2}=2-\sqrt{2}+3\sqrt{2}-2=2\sqrt{2}\)

câu b với câu c giải thích ra dùm e đc kh ạ?

a,\(\sqrt{\left(\sqrt{3}-1\right)^2}\) \(+\sqrt{\left(\sqrt{3}+1\right)^2}=2\sqrt{3}\)

b. \(\sqrt{\left(2\sqrt{5}+2\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}=3\sqrt{5}\)

c,\(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}=4\)

d.\(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-2\right)^2}=2\sqrt{2}\)

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.

$=\sqrt{2+2.3\sqrt{2}+3^2}-\sqrt{2-2.3\sqrt{2}+3^2}$

$=\sqrt{(\sqrt{2}+3)^2}-\sqrt{(\sqrt{2}-3)^2}$
$=|\sqrt{2}+3|-|\sqrt{2}-3|$

$=(\sqrt{2}+3)-(3-\sqrt{2})=2\sqrt{2}$

= \(3\sqrt{2}\)- 4 +  \(\sqrt{\left(2\sqrt{2}+3\right)^2}\)

= \(3\sqrt{2}\)-4 +   \(2\sqrt{2}\)+ 3

= \(5\sqrt{2}\)- 1

#mã mã#

Ta có: \(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)

\(=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)

\(=-13\sqrt{3}\)

\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\\ =2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}=-13\sqrt{3}\)

\(D=\sqrt{9+6\sqrt{2}}-\sqrt{9-6\sqrt{2}}-\sqrt{21-12\sqrt{3}}\)

\(D=\sqrt{9+2.\sqrt{3}.\sqrt{3}.\sqrt{2}}-\sqrt{9-2\sqrt{3}.\sqrt{3}.\sqrt{2}}-\sqrt{21-2.2\sqrt{3}.3}\)

\(D=\sqrt{\left(\sqrt{6}\right)^2+2\sqrt{6}.\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{6}\right)^2-2\sqrt{6}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)

\(-\sqrt{3^2-2.3.2\sqrt{3}+\left(2\sqrt{3}\right)^2}\)

\(D=\sqrt{\left(\sqrt{6}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{6}-\sqrt{3}\right)^2}-\sqrt{\left(3-2\sqrt{3}\right)^2}\)

\(D=\sqrt{6}+\sqrt{3}-\sqrt{6}+\sqrt{3}-2\sqrt{3}+3=3\)

1. \(=\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}+\sqrt{3+\sqrt{\left(2\sqrt{3}+1\right)^2}}=\sqrt{5-2\sqrt{3}-1}+\sqrt{3+2\sqrt{3}+1}=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)

1/ \(\sqrt{5-\sqrt{13+4\sqrt{3}}}+\sqrt{3+\sqrt{13+4\sqrt{3}}}\)

\(=\sqrt{5-\left(1+\sqrt{12}\right)^2}+\sqrt{3+\left(1+\sqrt{12}\right)^2}\)

\(=\sqrt{5-\left|1+\sqrt{12}\right|}+\sqrt{3+\left|1+\sqrt{12}\right|}\)

\(=\sqrt{5-1-\sqrt{12}}+\sqrt{3+1+\sqrt{12}}\)

\(=\sqrt{4-\sqrt{12}}+\sqrt{4+\sqrt{12}}\)

\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)

\(=\left|\sqrt{3}-1\right|+\left|\sqrt{3}+1\right|\)

\(=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)

2/ a) \(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)

\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)

\(=\left|\sqrt{3}-1\right|+\left|\sqrt{3}+1\right|=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)

b) \(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)

\(=\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(=\left|1+\sqrt{5}\right|-\left|\sqrt{5}-1\right|=1+\sqrt{5}-\sqrt{5}+1=2\)

c) \(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\)

\(=\sqrt{\left(\sqrt{9}-\sqrt{8}\right)^2}+\sqrt{\left(1+\sqrt{8}\right)^2}\)

\(=\left|\sqrt{9}-\sqrt{8}\right|+\left|1+\sqrt{8}\right|\)

\(=\sqrt{9}-\sqrt{8}+1+\sqrt{8}=3+1=4\)

Chúc các bạn học tốt nhá!