与えられた式 $(x+2y+1)(x-2y-1)$ を展開し、整理せよ。代数学展開因数分解多項式2025/5/291. 問題の内容与えられた式 (x+2y+1)(x−2y−1)(x+2y+1)(x-2y-1)(x+2y+1)(x−2y−1) を展開し、整理せよ。2. 解き方の手順与えられた式 (x+2y+1)(x−2y−1)(x+2y+1)(x-2y-1)(x+2y+1)(x−2y−1) を展開する。(x+2y+1)(x−2y−1)=x(x−2y−1)+2y(x−2y−1)+1(x−2y−1)(x+2y+1)(x-2y-1) = x(x-2y-1) + 2y(x-2y-1) + 1(x-2y-1)(x+2y+1)(x−2y−1)=x(x−2y−1)+2y(x−2y−1)+1(x−2y−1)=x2−2xy−x+2xy−4y2−2y+x−2y−1= x^2 - 2xy - x + 2xy - 4y^2 - 2y + x - 2y - 1=x2−2xy−x+2xy−4y2−2y+x−2y−1=x2−2xy−x+2xy−4y2−2y+x−2y−1= x^2 - 2xy - x + 2xy - 4y^2 - 2y + x - 2y - 1=x2−2xy−x+2xy−4y2−2y+x−2y−1=x2−4y2−4y−1= x^2 - 4y^2 - 4y - 1=x2−4y2−4y−1=x2−(4y2+4y+1)= x^2 - (4y^2 + 4y + 1)=x2−(4y2+4y+1)=x2−(2y+1)2= x^2 - (2y+1)^2=x2−(2y+1)2=x2−(2y+1)2=(x+(2y+1))(x−(2y+1))= x^2 - (2y+1)^2 = (x + (2y+1))(x - (2y+1))=x2−(2y+1)2=(x+(2y+1))(x−(2y+1))=(x+2y+1)(x−2y−1)= (x+2y+1)(x-2y-1)=(x+2y+1)(x−2y−1)または、以下のように考えることもできます。x+1=Ax+1=Ax+1=A とおくと、与式は(A+2y)(A−2y)=A2−(2y)2=A2−4y2(A+2y)(A-2y) = A^2 - (2y)^2 = A^2 - 4y^2(A+2y)(A−2y)=A2−(2y)2=A2−4y2A=x+1A=x+1A=x+1を代入して、(x+1)2−4y2=x2+2x+1−4y2=x2−4y2+2x+1(x+1)^2 - 4y^2 = x^2+2x+1-4y^2 = x^2 - 4y^2 + 2x + 1(x+1)2−4y2=x2+2x+1−4y2=x2−4y2+2x+1与式をもう一度確認すると (x+2y+1)(x−2y−1)(x+2y+1)(x-2y-1)(x+2y+1)(x−2y−1) です。これを (x+1+2y)(x+1−2y)(x+1+2y)(x+1-2y)(x+1+2y)(x+1−2y) と変形します。ここで、x+1=Ax+1 = Ax+1=A と置くと、(A+2y)(A−2y)=A2−(2y)2=A2−4y2(A+2y)(A-2y) = A^2 - (2y)^2 = A^2 - 4y^2(A+2y)(A−2y)=A2−(2y)2=A2−4y2A=x+1A=x+1A=x+1 を代入すると、(x+1)2−4y2=x2+2x+1−4y2=x2−4y2+2x+1(x+1)^2 - 4y^2 = x^2 + 2x + 1 - 4y^2 = x^2 - 4y^2 + 2x + 1(x+1)2−4y2=x2+2x+1−4y2=x2−4y2+2x+13. 最終的な答えx2−4y2+2x+1x^2 - 4y^2 + 2x + 1x2−4y2+2x+1