与えられた式 $(xy-1)(x-1)(y+1)-xy$ を展開し、整理して簡単にします。代数学式の展開多項式2025/4/211. 問題の内容与えられた式 (xy−1)(x−1)(y+1)−xy(xy-1)(x-1)(y+1)-xy(xy−1)(x−1)(y+1)−xy を展開し、整理して簡単にします。2. 解き方の手順まず、(xy−1)(x−1)(xy-1)(x-1)(xy−1)(x−1) を展開します。(xy−1)(x−1)=xy(x−1)−(x−1)=x2y−xy−x+1(xy-1)(x-1) = xy(x-1) - (x-1) = x^2y - xy - x + 1(xy−1)(x−1)=xy(x−1)−(x−1)=x2y−xy−x+1次に、この結果に (y+1)(y+1)(y+1) を掛けます。(x2y−xy−x+1)(y+1)=x2y(y+1)−xy(y+1)−x(y+1)+(y+1)(x^2y - xy - x + 1)(y+1) = x^2y(y+1) - xy(y+1) - x(y+1) + (y+1)(x2y−xy−x+1)(y+1)=x2y(y+1)−xy(y+1)−x(y+1)+(y+1)=x2y2+x2y−xy2−xy−xy−x+y+1= x^2y^2 + x^2y - xy^2 - xy - xy - x + y + 1=x2y2+x2y−xy2−xy−xy−x+y+1=x2y2+x2y−xy2−2xy−x+y+1= x^2y^2 + x^2y - xy^2 - 2xy - x + y + 1=x2y2+x2y−xy2−2xy−x+y+1最後に、この結果から xyxyxy を引きます。x2y2+x2y−xy2−2xy−x+y+1−xy=x2y2+x2y−xy2−3xy−x+y+1x^2y^2 + x^2y - xy^2 - 2xy - x + y + 1 - xy = x^2y^2 + x^2y - xy^2 - 3xy - x + y + 1x2y2+x2y−xy2−2xy−x+y+1−xy=x2y2+x2y−xy2−3xy−x+y+13. 最終的な答えx2y2+x2y−xy2−3xy−x+y+1x^2y^2 + x^2y - xy^2 - 3xy - x + y + 1x2y2+x2y−xy2−3xy−x+y+1