Bài 2:
a: Ta có: \(\sqrt{4-2\sqrt{3}}-\sqrt{3}\)
\(=\sqrt{3}-1-\sqrt{3}\)
=-1
b: Ta có: \(\sqrt{11-6\sqrt{2}}-3+\sqrt{2}\)
\(=3-\sqrt{2}-3+\sqrt{2}\)
=0
c: Ta có: \(\sqrt{9-4\sqrt{5}}-2\sqrt{5}\)
\(=\sqrt{5}-2-2\sqrt{5}\)
\(=-\sqrt{5}-2\)
d: Ta có: \(\sqrt{29+4\sqrt{7}}-\sqrt{11-4\sqrt{7}}\)
\(=2\sqrt{7}+1-2+\sqrt{7}\)
\(=3\sqrt{7}-1\)
Bài 1:
a) \(\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+\sqrt{1}\right)^2}=\sqrt{2}+1\)
b) \(\sqrt{28-6\sqrt{3}}=\sqrt{\left(\sqrt{27}-1\right)^2}=\sqrt{27}-1\)
c) \(\sqrt{4\sqrt{5}+9}=\sqrt{\left(\sqrt{5}-2\right)^2}=\sqrt{5}-2\)
d) \(\sqrt{29+4\sqrt{7}}=\sqrt{\left(\sqrt{28}+1\right)^2}=\sqrt{28}+1\)