画像にある数式を展開して簡単にします。代数学式の展開多項式2025/5/10はい、承知いたしました。画像にあるいくつかの数式を展開します。1. 問題の内容画像にある数式を展開して簡単にします。2. 解き方の手順各数式を分配法則を用いて展開します。(1) (b+c)(d+a)=b(d+a)+c(d+a)=bd+ba+cd+ca(b+c)(d+a) = b(d+a) + c(d+a) = bd + ba + cd + ca(b+c)(d+a)=b(d+a)+c(d+a)=bd+ba+cd+ca(2) (x−2)(y+3)=x(y+3)−2(y+3)=xy+3x−2y−6(x-2)(y+3) = x(y+3) - 2(y+3) = xy + 3x - 2y - 6(x−2)(y+3)=x(y+3)−2(y+3)=xy+3x−2y−6(3) (x+6)(x+4)=x(x+4)+6(x+4)=x2+4x+6x+24=x2+10x+24(x+6)(x+4) = x(x+4) + 6(x+4) = x^2 + 4x + 6x + 24 = x^2 + 10x + 24(x+6)(x+4)=x(x+4)+6(x+4)=x2+4x+6x+24=x2+10x+24(4) (x+5)(x−7)=x(x−7)+5(x−7)=x2−7x+5x−35=x2−2x−35(x+5)(x-7) = x(x-7) + 5(x-7) = x^2 - 7x + 5x - 35 = x^2 - 2x - 35(x+5)(x−7)=x(x−7)+5(x−7)=x2−7x+5x−35=x2−2x−35(5) (x+2y)(x−y)=x(x−y)+2y(x−y)=x2−xy+2xy−2y2=x2+xy−2y2(x+2y)(x-y) = x(x-y) + 2y(x-y) = x^2 - xy + 2xy - 2y^2 = x^2 + xy - 2y^2(x+2y)(x−y)=x(x−y)+2y(x−y)=x2−xy+2xy−2y2=x2+xy−2y2(6) (2x−3y)(3x−5y)=2x(3x−5y)−3y(3x−5y)=6x2−10xy−9xy+15y2=6x2−19xy+15y2(2x-3y)(3x-5y) = 2x(3x-5y) - 3y(3x-5y) = 6x^2 - 10xy - 9xy + 15y^2 = 6x^2 - 19xy + 15y^2(2x−3y)(3x−5y)=2x(3x−5y)−3y(3x−5y)=6x2−10xy−9xy+15y2=6x2−19xy+15y2(7) (2a2−a−2)(a+4)=2a2(a+4)−a(a+4)−2(a+4)=2a3+8a2−a2−4a−2a−8=2a3+7a2−6a−8(2a^2 - a - 2)(a+4) = 2a^2(a+4) - a(a+4) - 2(a+4) = 2a^3 + 8a^2 - a^2 - 4a - 2a - 8 = 2a^3 + 7a^2 - 6a - 8(2a2−a−2)(a+4)=2a2(a+4)−a(a+4)−2(a+4)=2a3+8a2−a2−4a−2a−8=2a3+7a2−6a−8(8) (−2a+3)(a2+2a−2)=−2a(a2+2a−2)+3(a2+2a−2)=−2a3−4a2+4a+3a2+6a−6=−2a3−a2+10a−6(-2a+3)(a^2+2a-2) = -2a(a^2+2a-2) + 3(a^2+2a-2) = -2a^3 - 4a^2 + 4a + 3a^2 + 6a - 6 = -2a^3 - a^2 + 10a - 6(−2a+3)(a2+2a−2)=−2a(a2+2a−2)+3(a2+2a−2)=−2a3−4a2+4a+3a2+6a−6=−2a3−a2+10a−63. 最終的な答え(1) bd+ba+cd+cabd + ba + cd + cabd+ba+cd+ca(2) xy+3x−2y−6xy + 3x - 2y - 6xy+3x−2y−6(3) x2+10x+24x^2 + 10x + 24x2+10x+24(4) x2−2x−35x^2 - 2x - 35x2−2x−35(5) x2+xy−2y2x^2 + xy - 2y^2x2+xy−2y2(6) 6x2−19xy+15y26x^2 - 19xy + 15y^26x2−19xy+15y2(7) 2a3+7a2−6a−82a^3 + 7a^2 - 6a - 82a3+7a2−6a−8(8) −2a3−a2+10a−6-2a^3 - a^2 + 10a - 6−2a3−a2+10a−6