次の等式を証明せよ。 $(a^2 + 2b^2)(c^2 + 2d^2) = (ac + 2bd)^2 + 2(ad - bc)^2$代数学等式の証明式の展開代数2025/7/221. 問題の内容次の等式を証明せよ。(a2+2b2)(c2+2d2)=(ac+2bd)2+2(ad−bc)2(a^2 + 2b^2)(c^2 + 2d^2) = (ac + 2bd)^2 + 2(ad - bc)^2(a2+2b2)(c2+2d2)=(ac+2bd)2+2(ad−bc)22. 解き方の手順まず、右辺を展開します。(ac+2bd)2+2(ad−bc)2=(a2c2+4acbd+4b2d2)+2(a2d2−2adbc+b2c2)(ac + 2bd)^2 + 2(ad - bc)^2 = (a^2c^2 + 4acbd + 4b^2d^2) + 2(a^2d^2 - 2adbc + b^2c^2)(ac+2bd)2+2(ad−bc)2=(a2c2+4acbd+4b2d2)+2(a2d2−2adbc+b2c2)=a2c2+4acbd+4b2d2+2a2d2−4adbc+2b2c2= a^2c^2 + 4acbd + 4b^2d^2 + 2a^2d^2 - 4adbc + 2b^2c^2=a2c2+4acbd+4b2d2+2a2d2−4adbc+2b2c2=a2c2+4b2d2+2a2d2+2b2c2= a^2c^2 + 4b^2d^2 + 2a^2d^2 + 2b^2c^2=a2c2+4b2d2+2a2d2+2b2c2次に、左辺を展開します。(a2+2b2)(c2+2d2)=a2c2+2a2d2+2b2c2+4b2d2(a^2 + 2b^2)(c^2 + 2d^2) = a^2c^2 + 2a^2d^2 + 2b^2c^2 + 4b^2d^2(a2+2b2)(c2+2d2)=a2c2+2a2d2+2b2c2+4b2d2右辺と左辺の展開結果が一致することを示します。a2c2+4b2d2+2a2d2+2b2c2=a2c2+2a2d2+2b2c2+4b2d2a^2c^2 + 4b^2d^2 + 2a^2d^2 + 2b^2c^2 = a^2c^2 + 2a^2d^2 + 2b^2c^2 + 4b^2d^2a2c2+4b2d2+2a2d2+2b2c2=a2c2+2a2d2+2b2c2+4b2d23. 最終的な答え(a2+2b2)(c2+2d2)=(ac+2bd)2+2(ad−bc)2(a^2 + 2b^2)(c^2 + 2d^2) = (ac + 2bd)^2 + 2(ad - bc)^2(a2+2b2)(c2+2d2)=(ac+2bd)2+2(ad−bc)2