与えられた数式 $x^2(y-1) + y^2(x-1) - 2xy + x + y$ を因数分解する問題です。代数学因数分解多項式展開共通因数2025/3/271. 問題の内容与えられた数式 x2(y−1)+y2(x−1)−2xy+x+yx^2(y-1) + y^2(x-1) - 2xy + x + yx2(y−1)+y2(x−1)−2xy+x+y を因数分解する問題です。2. 解き方の手順まず、数式を展開します。x2(y−1)+y2(x−1)−2xy+x+y=x2y−x2+xy2−y2−2xy+x+yx^2(y-1) + y^2(x-1) - 2xy + x + y = x^2y - x^2 + xy^2 - y^2 - 2xy + x + yx2(y−1)+y2(x−1)−2xy+x+y=x2y−x2+xy2−y2−2xy+x+y次に、同類項をまとめます。x2y−x2+xy2−y2−2xy+x+y=x2y+xy2−x2−y2−2xy+x+yx^2y - x^2 + xy^2 - y^2 - 2xy + x + y = x^2y + xy^2 - x^2 - y^2 - 2xy + x + yx2y−x2+xy2−y2−2xy+x+y=x2y+xy2−x2−y2−2xy+x+yさらに、式を整理して、因数分解しやすい形に並べ替えます。x2y+xy2−x2−y2−2xy+x+y=xy(x+y)−(x2+2xy+y2)+(x+y)x^2y + xy^2 - x^2 - y^2 - 2xy + x + y = xy(x+y) - (x^2 + 2xy + y^2) + (x+y)x2y+xy2−x2−y2−2xy+x+y=xy(x+y)−(x2+2xy+y2)+(x+y)ここで、x2+2xy+y2=(x+y)2x^2 + 2xy + y^2 = (x+y)^2x2+2xy+y2=(x+y)2 であることを利用します。xy(x+y)−(x2+2xy+y2)+(x+y)=xy(x+y)−(x+y)2+(x+y)xy(x+y) - (x^2 + 2xy + y^2) + (x+y) = xy(x+y) - (x+y)^2 + (x+y)xy(x+y)−(x2+2xy+y2)+(x+y)=xy(x+y)−(x+y)2+(x+y)(x+y)(x+y)(x+y)を共通因数としてくくり出します。xy(x+y)−(x+y)2+(x+y)=(x+y)(xy−(x+y)+1)xy(x+y) - (x+y)^2 + (x+y) = (x+y)(xy - (x+y) + 1)xy(x+y)−(x+y)2+(x+y)=(x+y)(xy−(x+y)+1)最後に、括弧の中身を整理します。(x+y)(xy−(x+y)+1)=(x+y)(xy−x−y+1)(x+y)(xy - (x+y) + 1) = (x+y)(xy - x - y + 1)(x+y)(xy−(x+y)+1)=(x+y)(xy−x−y+1)さらに、括弧の中身を因数分解します。(x+y)(xy−x−y+1)=(x+y)(x(y−1)−(y−1))=(x+y)(x−1)(y−1)(x+y)(xy - x - y + 1) = (x+y)(x(y-1) - (y-1)) = (x+y)(x-1)(y-1)(x+y)(xy−x−y+1)=(x+y)(x(y−1)−(y−1))=(x+y)(x−1)(y−1)3. 最終的な答え(x+y)(x−1)(y−1)(x+y)(x-1)(y-1)(x+y)(x−1)(y−1)