Question

Tính \(\sqrt{5-2\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)

GIÚP MÌNH VỚI CÁC BẠN ƠIII

 

Answer 1

\(\sqrt{5-2\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)

= \(\sqrt{5-2\sqrt{2+\sqrt{8+4\sqrt{2}+1}}}\)

= \(\sqrt{5-2\sqrt{2+\sqrt{\left(\sqrt{8}+1\right)^2}}}\)

= \(\sqrt{5-2\sqrt{2+2\sqrt{2}+1}}\)

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

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

= \(\sqrt{5-2\sqrt{2}-2}\)

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

= \(\sqrt{2}-1\)

Answer 2

CẢM ƠN BẠN RẤT NHIỀU <3

Answer 3

CẢM ƠN BẠN RẤT NHIỀU <3

Answer 4

\(\sqrt{5-2\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)

\(=\sqrt{5-2\sqrt{2+\sqrt{\left(2\sqrt{2}\right)^2+2\cdot2\sqrt{2}\cdot1+1}}}\)

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

\(=\sqrt{5-2\sqrt{2+2\sqrt{2}+1}}\)( vì \(2\sqrt{2}+1>0\))

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

\(=\sqrt{5-2\left(\sqrt{2}+1\right)}\)( vì \(\sqrt{2}+1>0\))

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

\(=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\)( vì \(\sqrt{2}-1>0\))