Lời giải:
\(A=\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)
\(A\sqrt{2}=\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5+2\sqrt{5}+1}+\sqrt{5-2\sqrt{5}+1}\)
\(=\sqrt{(\sqrt{5}+1)^2}+\sqrt{(\sqrt{5}-1)^2}\)
\(\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)
\(\Rightarrow A=\frac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)
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\(B=(3+\sqrt{5})(\sqrt{10}-\sqrt{2})\sqrt{3-\sqrt{5}}\)
\(=(3+\sqrt{5}(\sqrt{5}-1)\sqrt{6-2\sqrt{5}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1)\sqrt{5-2\sqrt{5}+1}=(3+\sqrt{5})(\sqrt{5}-1)\sqrt{(\sqrt{5}-1)^2}\)
\(=(3+\sqrt{5})(\sqrt{5}-1)(\sqrt{5}-1)=(3+\sqrt{5})(6-2\sqrt{5})\)
\(=(3+\sqrt{5}).2(3-\sqrt{5})=2(3^2-5)=8\)
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Phần C bạn xem đã chép đúng đề chưa vậy?
