与えられた6つの式を展開する問題です。代数学式の展開分配法則多項式2025/7/191. 問題の内容与えられた6つの式を展開する問題です。2. 解き方の手順各問題について、分配法則を用いて展開します。(1) (x−4)(y+2)(x-4)(y+2)(x−4)(y+2)x(y+2)−4(y+2)=xy+2x−4y−8x(y+2) - 4(y+2) = xy + 2x - 4y - 8x(y+2)−4(y+2)=xy+2x−4y−8(2) (a+c)(b−d)(a+c)(b-d)(a+c)(b−d)a(b−d)+c(b−d)=ab−ad+cb−cda(b-d) + c(b-d) = ab - ad + cb - cda(b−d)+c(b−d)=ab−ad+cb−cd(3) (x+5)(x+1)(x+5)(x+1)(x+5)(x+1)x(x+1)+5(x+1)=x2+x+5x+5=x2+6x+5x(x+1) + 5(x+1) = x^2 + x + 5x + 5 = x^2 + 6x + 5x(x+1)+5(x+1)=x2+x+5x+5=x2+6x+5(4) (4x−3)(x−6)(4x-3)(x-6)(4x−3)(x−6)4x(x−6)−3(x−6)=4x2−24x−3x+18=4x2−27x+184x(x-6) - 3(x-6) = 4x^2 - 24x - 3x + 18 = 4x^2 - 27x + 184x(x−6)−3(x−6)=4x2−24x−3x+18=4x2−27x+18(5) (a+7)(a−2b+4)(a+7)(a-2b+4)(a+7)(a−2b+4)a(a−2b+4)+7(a−2b+4)=a2−2ab+4a+7a−14b+28=a2−2ab+11a−14b+28a(a-2b+4) + 7(a-2b+4) = a^2 - 2ab + 4a + 7a - 14b + 28 = a^2 - 2ab + 11a - 14b + 28a(a−2b+4)+7(a−2b+4)=a2−2ab+4a+7a−14b+28=a2−2ab+11a−14b+28(6) (5x−y−2)(4x−3y)(5x-y-2)(4x-3y)(5x−y−2)(4x−3y)5x(4x−3y)−y(4x−3y)−2(4x−3y)=20x2−15xy−4xy+3y2−8x+6y=20x2−19xy+3y2−8x+6y5x(4x-3y) - y(4x-3y) - 2(4x-3y) = 20x^2 - 15xy - 4xy + 3y^2 - 8x + 6y = 20x^2 - 19xy + 3y^2 - 8x + 6y5x(4x−3y)−y(4x−3y)−2(4x−3y)=20x2−15xy−4xy+3y2−8x+6y=20x2−19xy+3y2−8x+6y3. 最終的な答え(1) xy+2x−4y−8xy + 2x - 4y - 8xy+2x−4y−8(2) ab−ad+cb−cdab - ad + cb - cdab−ad+cb−cd(3) x2+6x+5x^2 + 6x + 5x2+6x+5(4) 4x2−27x+184x^2 - 27x + 184x2−27x+18(5) a2−2ab+11a−14b+28a^2 - 2ab + 11a - 14b + 28a2−2ab+11a−14b+28(6) 20x2−19xy+3y2−8x+6y20x^2 - 19xy + 3y^2 - 8x + 6y20x2−19xy+3y2−8x+6y