次の式を展開しなさい。 $(3x+1)(2x-1)-6(x-1)^2$代数学式の展開多項式代数2025/5/181. 問題の内容次の式を展開しなさい。(3x+1)(2x−1)−6(x−1)2(3x+1)(2x-1)-6(x-1)^2(3x+1)(2x−1)−6(x−1)22. 解き方の手順まず、(3x+1)(2x−1)(3x+1)(2x-1)(3x+1)(2x−1) を展開します。(3x+1)(2x−1)=3x(2x)+3x(−1)+1(2x)+1(−1)=6x2−3x+2x−1=6x2−x−1(3x+1)(2x-1) = 3x(2x) + 3x(-1) + 1(2x) + 1(-1) = 6x^2 - 3x + 2x - 1 = 6x^2 - x - 1(3x+1)(2x−1)=3x(2x)+3x(−1)+1(2x)+1(−1)=6x2−3x+2x−1=6x2−x−1次に、(x−1)2(x-1)^2(x−1)2 を展開します。(x−1)2=(x−1)(x−1)=x2−x−x+1=x2−2x+1(x-1)^2 = (x-1)(x-1) = x^2 - x - x + 1 = x^2 - 2x + 1(x−1)2=(x−1)(x−1)=x2−x−x+1=x2−2x+1次に、−6(x−1)2-6(x-1)^2−6(x−1)2 を展開します。−6(x−1)2=−6(x2−2x+1)=−6x2+12x−6-6(x-1)^2 = -6(x^2 - 2x + 1) = -6x^2 + 12x - 6−6(x−1)2=−6(x2−2x+1)=−6x2+12x−6最後に、すべての項をまとめます。(3x+1)(2x−1)−6(x−1)2=(6x2−x−1)+(−6x2+12x−6)=6x2−x−1−6x2+12x−6=(6x2−6x2)+(−x+12x)+(−1−6)=0x2+11x−7=11x−7(3x+1)(2x-1)-6(x-1)^2 = (6x^2 - x - 1) + (-6x^2 + 12x - 6) = 6x^2 - x - 1 - 6x^2 + 12x - 6 = (6x^2 - 6x^2) + (-x + 12x) + (-1 - 6) = 0x^2 + 11x - 7 = 11x - 7(3x+1)(2x−1)−6(x−1)2=(6x2−x−1)+(−6x2+12x−6)=6x2−x−1−6x2+12x−6=(6x2−6x2)+(−x+12x)+(−1−6)=0x2+11x−7=11x−73. 最終的な答え11x−711x-711x−7