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

Ta có: \(\sqrt{29-2\sqrt{180}}-\sqrt{9+4\sqrt{5}}\)

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

\(=\sqrt{5}-5\)

\(=\sqrt{20-2.\sqrt{20.9}+9}-\sqrt{4+2.2\sqrt{5}+5}\)

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

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

\(=\sqrt{5}-5\)

\(\sqrt{\sqrt{5-\sqrt{3-\sqrt{29-2\cdot3\cdot2\sqrt{5}}}}}=\)\(\sqrt{\sqrt{5-\sqrt{3-2\sqrt{5}+3}}}=\sqrt{\sqrt{5-\sqrt{6-2\sqrt{5}}}}\)=\(\sqrt{\sqrt{5-\sqrt{5}+1}}=\sqrt{1}=1\)

Cần lắm hông

a/ \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

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

\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)

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

b,c tương tự

Biểu thức A chị tính A² rồi sẽ tính đc A

Biểu thức B ko bt có sai đề ở căn thứ 2 ko ạ

Nếu nhân B với căn 2 thì cái căn thức nhất tách đc thành hđt (a+b)² đấy ạ nhưng cái căn thứ 2 thì ko tách đc

đề câu B chả sai đi chỗ nào :)) tại tụi m tách sai thôi =)) 

\(B=\sqrt{29+6\sqrt{6}}-\sqrt{32-6\sqrt{15}}\)

\(B=\sqrt{\left(3\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(3\sqrt{3}-\sqrt{5}\right)^2}\) ( tách ra hằng đẳng thức ) 

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

\(B=\sqrt{2}+\sqrt{5}\)

nuột không :)) 

Ta có: \(\sqrt{29+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}+5\sqrt{2}\)

\(=\sqrt{29+30\sqrt{2+\sqrt{8+2\cdot2\sqrt{2}\cdot1+1}}}+5\sqrt{2}\)

\(=\sqrt{29+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}+5\sqrt{2}\)

\(=\sqrt{29+30\sqrt{2+2\sqrt{2}+1}}+5\sqrt{2}\)

\(=\sqrt{29+30\sqrt{2+2\sqrt{2}\cdot1+1}}+5\sqrt{2}\)

\(=\sqrt{29+30\sqrt{\left(\sqrt{2}+1\right)^2}}+5\sqrt{2}\)

\(=\sqrt{29+30\left(\sqrt{2}+1\right)}+5\sqrt{2}\)

\(=\sqrt{29+30\sqrt{2}+30}+5\sqrt{2}\)

\(=\sqrt{9+2\cdot3\cdot5\sqrt{2}+50}+5\sqrt{2}\)

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

\(=3+5\sqrt{2}+5\sqrt{2}=3+10\sqrt{2}\)

\(\sqrt{24+8\sqrt{5}}+\) \(\sqrt{9-4\sqrt{5}}=\) \(\sqrt{\left(2\sqrt{5}\right)^2+2.2\sqrt{5}.2+4}\) + \(\sqrt{5-2\sqrt{5}.2+4}\)

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

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\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\) = \(\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)= \(\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)

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\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)

= \(\sqrt{13+30\sqrt{3+2\sqrt{2}}}=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{43+30\sqrt{2}}\) \(=\sqrt{\left(3\sqrt{2}+5\right)^2}=3\sqrt{2}+5\)

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\(a,=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\) \(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2.3\sqrt{20}+9}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\)\(=\sqrt{\sqrt{5}-\sqrt{6-\sqrt{20}}}\)

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

\(=\sqrt{1}=1\)

\(b,=\sqrt{3+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\) \(=\sqrt{3+30\sqrt{2+\sqrt{8+2\sqrt{8}+1}}}\)

\(=\sqrt{3+30\sqrt{2+\sqrt{\left(\sqrt{8}+1\right)^2}}}\)\(=\sqrt{3+30\sqrt{3+\sqrt{8}}}=\sqrt{3+30\sqrt{2+2\sqrt{2}+1}}\)

\(=\sqrt{3+30\sqrt{\left(\sqrt{2}+1\right)^2}}=\sqrt{3+30\sqrt{2}+30}=\sqrt{33+30\sqrt{2}}\)

a) Ta có: \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

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

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

\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)

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

=1

b) Ta có: \(\sqrt{3+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)

\(=\sqrt{3+30\sqrt{2+2\sqrt{2}+1}}\)

\(=\sqrt{3+30\left(\sqrt{2}+1\right)}\)

\(=\sqrt{33+30\sqrt{2}}\)