分式有理化得\((\sqrt2-1)^2\)，開方后得 \(\sqrt2-1\)，
於是 \(R = {{4r}\over {(2-\sqrt2)^2}}\) = \({2r} \over {3-2\sqrt2}\)
於是 \(R = {{4r}\over {(2-\sqrt2)^2}}\) = \({2r} \over {3-2\sqrt2}\)
锟斤拷锟洁辑时锟斤拷: 2021-05-04 03:19:22
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