\(=\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{3}}{3}\cdot\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{2}}{2}+\dfrac{3}{6}=\dfrac{\sqrt{2}}{2}+\dfrac{1}{2}=\dfrac{\sqrt{2}+1}{2}\left(B\right)\)
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