【5】次の式を展開しなさい。 (1) $(a+b-1)(a+b+1)$ (2) $(x-y-9)(x-y-2)$ (3) $(x+y-5)^2$代数学展開多項式因数分解置換2025/5/1はい、承知いたしました。写真の問題のうち、【5】の(1)から(3)までを解きます。1. 問題の内容【5】次の式を展開しなさい。(1) (a+b−1)(a+b+1)(a+b-1)(a+b+1)(a+b−1)(a+b+1)(2) (x−y−9)(x−y−2)(x-y-9)(x-y-2)(x−y−9)(x−y−2)(3) (x+y−5)2(x+y-5)^2(x+y−5)22. 解き方の手順(1) (a+b−1)(a+b+1)(a+b-1)(a+b+1)(a+b−1)(a+b+1)a+b=Aa+b = Aa+b=A と置換します。(A−1)(A+1)=A2−1(A-1)(A+1) = A^2 - 1(A−1)(A+1)=A2−1AAA を a+ba+ba+b に戻します。(a+b)2−1=a2+2ab+b2−1(a+b)^2 - 1 = a^2 + 2ab + b^2 - 1(a+b)2−1=a2+2ab+b2−1(2) (x−y−9)(x−y−2)(x-y-9)(x-y-2)(x−y−9)(x−y−2)x−y=Ax-y = Ax−y=A と置換します。(A−9)(A−2)=A2−11A+18(A-9)(A-2) = A^2 - 11A + 18(A−9)(A−2)=A2−11A+18AAA を x−yx-yx−y に戻します。(x−y)2−11(x−y)+18=x2−2xy+y2−11x+11y+18(x-y)^2 - 11(x-y) + 18 = x^2 - 2xy + y^2 - 11x + 11y + 18(x−y)2−11(x−y)+18=x2−2xy+y2−11x+11y+18(3) (x+y−5)2(x+y-5)^2(x+y−5)2(x+y−5)(x+y−5)(x+y-5)(x+y-5)(x+y−5)(x+y−5)=x(x+y−5)+y(x+y−5)−5(x+y−5)= x(x+y-5) + y(x+y-5) - 5(x+y-5)=x(x+y−5)+y(x+y−5)−5(x+y−5)=x2+xy−5x+yx+y2−5y−5x−5y+25= x^2 + xy - 5x + yx + y^2 - 5y - 5x - 5y + 25=x2+xy−5x+yx+y2−5y−5x−5y+25=x2+y2+2xy−10x−10y+25= x^2 + y^2 + 2xy - 10x - 10y + 25=x2+y2+2xy−10x−10y+253. 最終的な答え(1) a2+2ab+b2−1a^2 + 2ab + b^2 - 1a2+2ab+b2−1(2) x2−2xy+y2−11x+11y+18x^2 - 2xy + y^2 - 11x + 11y + 18x2−2xy+y2−11x+11y+18(3) x2+y2+2xy−10x−10y+25x^2 + y^2 + 2xy - 10x - 10y + 25x2+y2+2xy−10x−10y+25