Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6+3\sqrt{3}}\)
=\(\sqrt{21-2\sqrt{54}}+\sqrt{6+2\sqrt{18}+3}-\sqrt{4\cdot\left(6+3\sqrt{3}\right)}\)
=\(\sqrt{18-2\sqrt{54}+3}+\sqrt{6+2\sqrt{18}+3}-\sqrt{24+12\sqrt{3}}\)
=\(\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{6}+\sqrt{3}\right)^2}-\sqrt{24+2\sqrt{108}}\)
=\(\left|\sqrt{18}-\sqrt{3}\right|+\left|\sqrt{6}+\sqrt{3}\right|-\sqrt{\sqrt{18}+2\sqrt{108}+\sqrt{6}}\)
=\(\left|\sqrt{18}-\sqrt{3}\right|+\left|\sqrt{6}+\sqrt{3}\right|-\sqrt{\left(\sqrt{18}+\sqrt{6}\right)^2}\)
=\(\left|\sqrt{18}-\sqrt{3}\right|+\left|\sqrt{6}+\sqrt{3}\right|-\left|\sqrt{18}-\sqrt{6}\right|\)
=\(\sqrt{18}-\sqrt{3}+\sqrt{6}+\sqrt{3}-\sqrt{18}-\sqrt{6}\)
= 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
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: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)
\(\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6+3\sqrt{3}}\)
\(=3\sqrt{2}-\sqrt{3}+\sqrt{3}+\sqrt{6}-3\sqrt{2}-\sqrt{6}=0\)