Câu hỏi của Nguyễn Minh Anh

Question

Tính :$R=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}$

Nguyễn Lê Phước Thịnh · Accepted Answer

$=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\sqrt{4-2-\sqrt{2+\sqrt{3}}}$$=\sqrt{2+\sqrt{3}}\cdot\sqrt{4-2-\sqrt{3}}$=1Câu trả lời: $=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\sqrt{4-2-\sqrt{2+\sqrt{3}}}$$=\sqrt{2+\sqrt{3}}\cdot\sqrt{4-2-\sqrt{3}}$=1

Nguyễn Gia Phúc · Answer

=√2+√3⋅√2+√2+√3⋅√22−(2+√2+√3)2=2+3⋅2+2+3⋅22−(2+2+3)2=√2+√3⋅√2+√2+√3⋅√4−2−√2+√3=2+3⋅2+2+3⋅4−2−2+3=√2+√3⋅√2+√2+√3⋅√2−√2+√3=2+3⋅2+2+3⋅2−2+3=√2+√3⋅√4−2−√3=2+3⋅4−2−3=√2+√3⋅√2−√3=√4−3=1Câu trả lời: =√2+√3⋅√2+√2+√3⋅√22−(2+√2+√3)2=2+3⋅2+2+3⋅22−(2+2+3)2=√2+√3⋅√2+√2+√3⋅√4−2−√2+√3=2+3⋅2+2+3⋅4−2−2+3=√2+√3⋅√2+√2+√3⋅√2−√2+√3=2+3⋅2+2+3⋅2−2+3=√2+√3⋅√4−2−√3=2+3⋅4−2−3=√2+√3⋅√2−√3=√4−3=1

Nguyễn Gia Phúc · Answer

Câu trả lời:

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