与えられた8つの計算問題を解く。代数学式の展開因数分解多項式計算2025/6/241. 問題の内容与えられた8つの計算問題を解く。2. 解き方の手順(1) −3a(4a+b)=−12a2−3ab-3a(4a + b) = -12a^2 - 3ab−3a(4a+b)=−12a2−3ab(2) (9x2+6xy)÷35x=(9x2+6xy)×53x=45x2+30xy3x=15x+10y(9x^2 + 6xy) \div \frac{3}{5}x = (9x^2 + 6xy) \times \frac{5}{3x} = \frac{45x^2 + 30xy}{3x} = 15x + 10y(9x2+6xy)÷53x=(9x2+6xy)×3x5=3x45x2+30xy=15x+10y(3) (x+3)(x−2)=x2−2x+3x−6=x2+x−6(x+3)(x-2) = x^2 - 2x + 3x - 6 = x^2 + x - 6(x+3)(x−2)=x2−2x+3x−6=x2+x−6(4) (x−y)(x+2y)=x2+2xy−xy−2y2=x2+xy−2y2(x-y)(x+2y) = x^2 + 2xy - xy - 2y^2 = x^2 + xy - 2y^2(x−y)(x+2y)=x2+2xy−xy−2y2=x2+xy−2y2(5) (x+3)2=x2+2(x)(3)+32=x2+6x+9(x+3)^2 = x^2 + 2(x)(3) + 3^2 = x^2 + 6x + 9(x+3)2=x2+2(x)(3)+32=x2+6x+9(6) (−x+y)2=(−x)2+2(−x)(y)+y2=x2−2xy+y2(-x+y)^2 = (-x)^2 + 2(-x)(y) + y^2 = x^2 - 2xy + y^2(−x+y)2=(−x)2+2(−x)(y)+y2=x2−2xy+y2(7) (2a+3b)(2a−3b)=(2a)2−(3b)2=4a2−9b2(2a+3b)(2a-3b) = (2a)^2 - (3b)^2 = 4a^2 - 9b^2(2a+3b)(2a−3b)=(2a)2−(3b)2=4a2−9b2(8) (a+b)(a+b+4)=a(a+b+4)+b(a+b+4)=a2+ab+4a+ab+b2+4b=a2+b2+2ab+4a+4b(a+b)(a+b+4) = a(a+b+4) + b(a+b+4) = a^2 + ab + 4a + ab + b^2 + 4b = a^2 + b^2 + 2ab + 4a + 4b(a+b)(a+b+4)=a(a+b+4)+b(a+b+4)=a2+ab+4a+ab+b2+4b=a2+b2+2ab+4a+4b3. 最終的な答え(1) −12a2−3ab-12a^2 - 3ab−12a2−3ab(2) 15x+10y15x + 10y15x+10y(3) x2+x−6x^2 + x - 6x2+x−6(4) x2+xy−2y2x^2 + xy - 2y^2x2+xy−2y2(5) x2+6x+9x^2 + 6x + 9x2+6x+9(6) x2−2xy+y2x^2 - 2xy + y^2x2−2xy+y2(7) 4a2−9b24a^2 - 9b^24a2−9b2(8) a2+b2+2ab+4a+4ba^2 + b^2 + 2ab + 4a + 4ba2+b2+2ab+4a+4b