与えられた式を計算します。 式は $\frac{1}{1-\sqrt{2}} - \frac{1}{\sqrt{2}-\sqrt{3}} + \frac{1}{\sqrt{3}-2}$ です。代数学式の計算有理化平方根2025/6/11. 問題の内容与えられた式を計算します。式は 11−2−12−3+13−2\frac{1}{1-\sqrt{2}} - \frac{1}{\sqrt{2}-\sqrt{3}} + \frac{1}{\sqrt{3}-2}1−21−2−31+3−21 です。2. 解き方の手順各項の分母を有理化します。11−2=11−2⋅1+21+2=1+21−2=1+2−1=−1−2\frac{1}{1-\sqrt{2}} = \frac{1}{1-\sqrt{2}} \cdot \frac{1+\sqrt{2}}{1+\sqrt{2}} = \frac{1+\sqrt{2}}{1-2} = \frac{1+\sqrt{2}}{-1} = -1-\sqrt{2}1−21=1−21⋅1+21+2=1−21+2=−11+2=−1−212−3=12−3⋅2+32+3=2+32−3=2+3−1=−2−3\frac{1}{\sqrt{2}-\sqrt{3}} = \frac{1}{\sqrt{2}-\sqrt{3}} \cdot \frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}+\sqrt{3}} = \frac{\sqrt{2}+\sqrt{3}}{2-3} = \frac{\sqrt{2}+\sqrt{3}}{-1} = -\sqrt{2}-\sqrt{3}2−31=2−31⋅2+32+3=2−32+3=−12+3=−2−313−2=13−2⋅3+23+2=3+23−4=3+2−1=−3−2\frac{1}{\sqrt{3}-2} = \frac{1}{\sqrt{3}-2} \cdot \frac{\sqrt{3}+2}{\sqrt{3}+2} = \frac{\sqrt{3}+2}{3-4} = \frac{\sqrt{3}+2}{-1} = -\sqrt{3}-23−21=3−21⋅3+23+2=3−43+2=−13+2=−3−2与えられた式に代入します。(−1−2)−(−2−3)+(−3−2)=−1−2+2+3−3−2=−1−2=−3(-1-\sqrt{2}) - (-\sqrt{2}-\sqrt{3}) + (-\sqrt{3}-2) = -1-\sqrt{2} + \sqrt{2} + \sqrt{3} - \sqrt{3} - 2 = -1-2 = -3(−1−2)−(−2−3)+(−3−2)=−1−2+2+3−3−2=−1−2=−33. 最終的な答え-3