与えられた式 $(x+1)(y+1)(xy+1)+xy$ を展開し、整理せよ。代数学式の展開多項式2025/4/291. 問題の内容与えられた式 (x+1)(y+1)(xy+1)+xy(x+1)(y+1)(xy+1)+xy(x+1)(y+1)(xy+1)+xy を展開し、整理せよ。2. 解き方の手順まず、(x+1)(y+1)(x+1)(y+1)(x+1)(y+1) を展開します。(x+1)(y+1)=xy+x+y+1(x+1)(y+1) = xy + x + y + 1(x+1)(y+1)=xy+x+y+1次に、上記の式と (xy+1)(xy+1)(xy+1) をかけます。(xy+x+y+1)(xy+1)=(xy)2+xy+x(xy)+x+y(xy)+y+xy+1=x2y2+xy+x2y+x+xy2+y+xy+1=x2y2+x2y+xy2+2xy+x+y+1(xy + x + y + 1)(xy + 1) = (xy)^2 + xy + x(xy) + x + y(xy) + y + xy + 1 = x^2y^2 + xy + x^2y + x + xy^2 + y + xy + 1 = x^2y^2 + x^2y + xy^2 + 2xy + x + y + 1(xy+x+y+1)(xy+1)=(xy)2+xy+x(xy)+x+y(xy)+y+xy+1=x2y2+xy+x2y+x+xy2+y+xy+1=x2y2+x2y+xy2+2xy+x+y+1最後に、上記の式に xyxyxy を加えます。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