$(1-\sqrt{2})(3+4\sqrt{2})$ を計算せよ。代数学式の計算平方根展開2025/6/81. 問題の内容(1−2)(3+42)(1-\sqrt{2})(3+4\sqrt{2})(1−2)(3+42) を計算せよ。2. 解き方の手順分配法則を用いて展開します。(a+b)(c+d)=ac+ad+bc+bd(a+b)(c+d) = ac + ad + bc + bd(a+b)(c+d)=ac+ad+bc+bd を利用します。1×3=31 \times 3 = 31×3=31×42=421 \times 4\sqrt{2} = 4\sqrt{2}1×42=42−2×3=−32-\sqrt{2} \times 3 = -3\sqrt{2}−2×3=−32−2×42=−4×2=−8-\sqrt{2} \times 4\sqrt{2} = -4 \times 2 = -8−2×42=−4×2=−8よって、(1−2)(3+42)=3+42−32−8(1-\sqrt{2})(3+4\sqrt{2}) = 3 + 4\sqrt{2} - 3\sqrt{2} - 8(1−2)(3+42)=3+42−32−8=(3−8)+(42−32)= (3-8) + (4\sqrt{2}-3\sqrt{2})=(3−8)+(42−32)=−5+2= -5 + \sqrt{2}=−5+23. 最終的な答え−5+2-5 + \sqrt{2}−5+2