$\frac{\sqrt{6}}{\sqrt{-12}}$ を計算せよ。代数学複素数根号計算2025/6/111. 問題の内容6−12\frac{\sqrt{6}}{\sqrt{-12}}−126 を計算せよ。2. 解き方の手順まず、−12\sqrt{-12}−12 を虚数単位 iii を用いて書き換えます。−12=12⋅−1=4⋅3⋅i=23i\sqrt{-12} = \sqrt{12} \cdot \sqrt{-1} = \sqrt{4 \cdot 3} \cdot i = 2\sqrt{3}i−12=12⋅−1=4⋅3⋅i=23iしたがって、6−12=623i\frac{\sqrt{6}}{\sqrt{-12}} = \frac{\sqrt{6}}{2\sqrt{3}i}−126=23i6分母と分子に 3\sqrt{3}3 をかけると、623i=6323i⋅3=182⋅3⋅i=9⋅26i=326i=22i\frac{\sqrt{6}}{2\sqrt{3}i} = \frac{\sqrt{6}\sqrt{3}}{2\sqrt{3}i \cdot \sqrt{3}} = \frac{\sqrt{18}}{2 \cdot 3 \cdot i} = \frac{\sqrt{9 \cdot 2}}{6i} = \frac{3\sqrt{2}}{6i} = \frac{\sqrt{2}}{2i}23i6=23i⋅363=2⋅3⋅i18=6i9⋅2=6i32=2i2分母と分子に iii をかけると、22i=2⋅i2i⋅i=2i2(−1)=−22i\frac{\sqrt{2}}{2i} = \frac{\sqrt{2} \cdot i}{2i \cdot i} = \frac{\sqrt{2}i}{2(-1)} = -\frac{\sqrt{2}}{2}i2i2=2i⋅i2⋅i=2(−1)2i=−22i3. 最終的な答え−22i-\frac{\sqrt{2}}{2}i−22i