\(\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6+3\sqrt{3}}\)

=\(\sqrt{21-2\sqrt{54}}+\sqrt{6+2\sqrt{18}+3}-\sqrt{4\cdot\left(6+3\sqrt{3}\right)}\)

=\(\sqrt{18-2\sqrt{54}+3}+\sqrt{6+2\sqrt{18}+3}-\sqrt{24+12\sqrt{3}}\)

=\(\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{6}+\sqrt{3}\right)^2}-\sqrt{24+2\sqrt{108}}\)

=\(\left|\sqrt{18}-\sqrt{3}\right|+\left|\sqrt{6}+\sqrt{3}\right|-\sqrt{\sqrt{18}+2\sqrt{108}+\sqrt{6}}\)

=\(\left|\sqrt{18}-\sqrt{3}\right|+\left|\sqrt{6}+\sqrt{3}\right|-\sqrt{\left(\sqrt{18}+\sqrt{6}\right)^2}\)

=\(\left|\sqrt{18}-\sqrt{3}\right|+\left|\sqrt{6}+\sqrt{3}\right|-\left|\sqrt{18}-\sqrt{6}\right|\)

=\(\sqrt{18}-\sqrt{3}+\sqrt{6}+\sqrt{3}-\sqrt{18}-\sqrt{6}\)

= 0

Hic câu dưới bị giải nhầm nha bạn :<

\(\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6+3\sqrt{3}}\)

=\(\sqrt{21-2\sqrt{54}}+\sqrt{6+2\sqrt{18}+3}-\sqrt{24+12\sqrt{3}}\)

=\(\sqrt{18-2\sqrt{54}+3}+\sqrt{6+2\sqrt{18}+3}-\sqrt{24+2\sqrt{108}}\)

=\(\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{6}+\sqrt{3}\right)^2}-\sqrt{18+2\sqrt{108}+6}\)

=\(\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{6}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{18}+\sqrt{6}\right)^2}\)

=\(\left|\sqrt{18}-\sqrt{3}\right|+\left|\sqrt{6}+\sqrt{3}\right|-\left|\sqrt{18}+\sqrt{6}\right|\)

=\(\sqrt{18}-\sqrt{3}+\sqrt{6}+\sqrt{3}-\sqrt{18}-\sqrt{6}\)

=0

0 nhé bạn, thực ra thì tui bấm máy tính, chớ tui ms hc lớp 7 hà,

tích vs nhé

nếu bấm máy thì chắc chị giải đc rồi em :> 

a)Đặt A=.......

Bình phương 2 vế rồi làm binh thường

1: \(=\sqrt{36}=6\)

2: \(=\sqrt{\left(15-9\right)\left(15+9\right)}=\sqrt{24\cdot6}=12\)

3: \(=3\sqrt{5}-1-3\sqrt{5}-1=-2\)

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

5: \(=\left(2+\sqrt{5}\right)\left(\sqrt{5}-2\right)=5-4=1\)

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

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

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

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

6) \(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\sqrt{7}-\sqrt{3}\)

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

a: \(=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)

b: \(=\left(-\sqrt{5}-2+\sqrt{5}-\sqrt{3}\right)\cdot\left(2\sqrt{3}+3\right)\)

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

\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)

c: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)