\(A=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\\ \Rightarrow A^2=2-\sqrt{3}+2\sqrt{2-\sqrt{3}}\sqrt{2+\sqrt{3}}+2+\sqrt{3}\\ \Rightarrow A^2=4+2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\\ \Rightarrow A^2=4+2\sqrt{2^2-\sqrt{3^2}}\\ \Rightarrow A^2=4+2\sqrt{1}\\ \Rightarrow A^2=6\\ \Rightarrow A=\pm\sqrt{6}\)
Mà \(A>0\Rightarrow A=\sqrt{6}\)
Lời giải:
\(B\sqrt{2}=\sqrt{8+2\sqrt{15}}-\sqrt{6-2\sqrt{5}}=\sqrt{5+2\sqrt{5}.\sqrt{3}+3}-\sqrt{5-2\sqrt{5}+1}\)
\(=\sqrt{(\sqrt{5}+\sqrt{3})^2}-\sqrt{(\sqrt{5}-1)^2}=|\sqrt{5}+\sqrt{3}|-|\sqrt{5}-1|\)
\(=\sqrt{5}+\sqrt{3}-(\sqrt{5}-1)=\sqrt{3}+1\)
$\Rightarrow B=\frac{\sqrt{3}+1}{\sqrt{2}}$
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\(C\sqrt{2}=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}-2\)
\(=\sqrt{(\sqrt{5}+1)^2}-\sqrt{(\sqrt{5}-1)^2}-2=|\sqrt{5}+1|-|\sqrt{5}-1|-2\)
\(=(\sqrt{5}+1)-(\sqrt{5}-1)-2=0\Rightarrow C=0\)
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\(D\sqrt{2}=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}+\sqrt{14}\)
\(=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}+\sqrt{14}=|\sqrt{7}-1|-|\sqrt{7}+1|+\sqrt{14}\)
\(=\sqrt{7}-1-(\sqrt{7}+1)+\sqrt{14}=-2+\sqrt{14}\)
\(\Rightarrow D=-\sqrt{2}+\sqrt{7}\)
giúp mình câu c bài 2
