a, Ta có : \(A=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)

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

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

\(\Rightarrow A=-\sqrt{2}\)

b, Ta có : \(B=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)

\(\Rightarrow B\sqrt{2}=\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2\)

\(=\sqrt{5+2\sqrt{5}+1}+\sqrt{9-2.3\sqrt{5}+5}-2\)

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

\(\Rightarrow B=\sqrt{2}\)

Lời giải:

\(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{(\sqrt{3}+1)^2}}}\)

\(=\sqrt{6+2\sqrt{2}.\sqrt{3-(\sqrt{3}+1)}}\)

\(=\sqrt{6+2\sqrt{2}.\sqrt{2-\sqrt{3}}}=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)

\(=\sqrt{6+2\sqrt{(\sqrt{3}-1)^2}}=\sqrt{6+2(\sqrt{3}-1)}=\sqrt{4+2\sqrt{3}}\)

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

\(=\sqrt{\left(2-\sqrt{3}\right)^2\left(26+15\sqrt{3}\right)}-\sqrt{\left(2+\sqrt{3}\right)^2\left(26-15\sqrt{3}\right)}=\)

\(=\sqrt{\left(7-4\sqrt{3}\right)\left(26+15\sqrt{3}\right)}-\sqrt{\left(7+4\sqrt{3}\right)\left(26-15\sqrt{3}\right)=}\)

\(=\sqrt{7.26+7.15\sqrt{3}-4.26\sqrt{3}-180}-\sqrt{7.26-7.15\sqrt{3}+4.26\sqrt{3}-180}=\)

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

Ta có: \(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)

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

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

TK:"https://www.google.com.vn/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&cad=rja&uact=8&ved=2ahUKEwjF-cL3pJzxAhWbPXAKHTjTAxwQFjABegQIBRAD&url=https%3A%2F%2Fhoidap247.com%2Fcau-hoi%2F996088&usg=AOvVaw3JxumatFPaPIuCWni48U22"

`sqrt{3-2sqrt2}-sqrt{3+2sqrt2}`

`=sqrt{2-2sqrt2+1}-sqrt{2+2sqrt2+1}`

`=sqrt{(sqrt2-1)^2}-sqrt{(sqrt2+1)^2}`

`=sqrt2-1-sqrt2-1=-2`

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

   = \(\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{3+2\sqrt{2}+2\sqrt{3}+2\sqrt{6}+3}\)

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

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

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

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

   = \(-\sqrt{3}\)

\(\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)

\(=\dfrac{\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.1+1^2}}{\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}}-\dfrac{\sqrt{\left(\sqrt{2}\right)^2+2.\sqrt{2}.1+1^2}}{\sqrt{3^2+2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}}\)

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

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

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

chịu,,, chắc toàn dấu cộng chứ tự nhiên có dấu trừ sao làm

nếu là dấu cộng cx khó đấy