与えられた整式 $A$ と $B$ について、$A+B$、$A-B$、$2A-3B$ を計算する問題です。代数学式の計算多項式加減算2025/4/151. 問題の内容与えられた整式 AAA と BBB について、A+BA+BA+B、A−BA-BA−B、2A−3B2A-3B2A−3B を計算する問題です。2. 解き方の手順(1) A=x2+3x+1A = x^2 + 3x + 1A=x2+3x+1、B=2x2−5x−1B = 2x^2 - 5x - 1B=2x2−5x−1 の場合* A+BA + BA+B の計算: A+B=(x2+3x+1)+(2x2−5x−1)=x2+2x2+3x−5x+1−1=3x2−2xA + B = (x^2 + 3x + 1) + (2x^2 - 5x - 1) = x^2 + 2x^2 + 3x - 5x + 1 - 1 = 3x^2 - 2xA+B=(x2+3x+1)+(2x2−5x−1)=x2+2x2+3x−5x+1−1=3x2−2x* A−BA - BA−B の計算: A−B=(x2+3x+1)−(2x2−5x−1)=x2−2x2+3x+5x+1+1=−x2+8x+2A - B = (x^2 + 3x + 1) - (2x^2 - 5x - 1) = x^2 - 2x^2 + 3x + 5x + 1 + 1 = -x^2 + 8x + 2A−B=(x2+3x+1)−(2x2−5x−1)=x2−2x2+3x+5x+1+1=−x2+8x+2* 2A−3B2A - 3B2A−3B の計算: 2A=2(x2+3x+1)=2x2+6x+22A = 2(x^2 + 3x + 1) = 2x^2 + 6x + 22A=2(x2+3x+1)=2x2+6x+2 3B=3(2x2−5x−1)=6x2−15x−33B = 3(2x^2 - 5x - 1) = 6x^2 - 15x - 33B=3(2x2−5x−1)=6x2−15x−3 2A−3B=(2x2+6x+2)−(6x2−15x−3)=2x2−6x2+6x+15x+2+3=−4x2+21x+52A - 3B = (2x^2 + 6x + 2) - (6x^2 - 15x - 3) = 2x^2 - 6x^2 + 6x + 15x + 2 + 3 = -4x^2 + 21x + 52A−3B=(2x2+6x+2)−(6x2−15x−3)=2x2−6x2+6x+15x+2+3=−4x2+21x+5(2) A=3x2−4xy+y2A = 3x^2 - 4xy + y^2A=3x2−4xy+y2、B=−2x2+xy−3y2B = -2x^2 + xy - 3y^2B=−2x2+xy−3y2 の場合* A+BA + BA+B の計算: A+B=(3x2−4xy+y2)+(−2x2+xy−3y2)=3x2−2x2−4xy+xy+y2−3y2=x2−3xy−2y2A + B = (3x^2 - 4xy + y^2) + (-2x^2 + xy - 3y^2) = 3x^2 - 2x^2 - 4xy + xy + y^2 - 3y^2 = x^2 - 3xy - 2y^2A+B=(3x2−4xy+y2)+(−2x2+xy−3y2)=3x2−2x2−4xy+xy+y2−3y2=x2−3xy−2y2* A−BA - BA−B の計算: A−B=(3x2−4xy+y2)−(−2x2+xy−3y2)=3x2+2x2−4xy−xy+y2+3y2=5x2−5xy+4y2A - B = (3x^2 - 4xy + y^2) - (-2x^2 + xy - 3y^2) = 3x^2 + 2x^2 - 4xy - xy + y^2 + 3y^2 = 5x^2 - 5xy + 4y^2A−B=(3x2−4xy+y2)−(−2x2+xy−3y2)=3x2+2x2−4xy−xy+y2+3y2=5x2−5xy+4y2* 2A−3B2A - 3B2A−3B の計算: 2A=2(3x2−4xy+y2)=6x2−8xy+2y22A = 2(3x^2 - 4xy + y^2) = 6x^2 - 8xy + 2y^22A=2(3x2−4xy+y2)=6x2−8xy+2y2 3B=3(−2x2+xy−3y2)=−6x2+3xy−9y23B = 3(-2x^2 + xy - 3y^2) = -6x^2 + 3xy - 9y^23B=3(−2x2+xy−3y2)=−6x2+3xy−9y2 2A−3B=(6x2−8xy+2y2)−(−6x2+3xy−9y2)=6x2+6x2−8xy−3xy+2y2+9y2=12x2−11xy+11y22A - 3B = (6x^2 - 8xy + 2y^2) - (-6x^2 + 3xy - 9y^2) = 6x^2 + 6x^2 - 8xy - 3xy + 2y^2 + 9y^2 = 12x^2 - 11xy + 11y^22A−3B=(6x2−8xy+2y2)−(−6x2+3xy−9y2)=6x2+6x2−8xy−3xy+2y2+9y2=12x2−11xy+11y2(3) A=2y2+5xy+x2A = 2y^2 + 5xy + x^2A=2y2+5xy+x2、B=−3xy+3x2−y2B = -3xy + 3x^2 - y^2B=−3xy+3x2−y2 の場合* A+BA + BA+B の計算: A+B=(2y2+5xy+x2)+(−3xy+3x2−y2)=x2+3x2+5xy−3xy+2y2−y2=4x2+2xy+y2A + B = (2y^2 + 5xy + x^2) + (-3xy + 3x^2 - y^2) = x^2 + 3x^2 + 5xy - 3xy + 2y^2 - y^2 = 4x^2 + 2xy + y^2A+B=(2y2+5xy+x2)+(−3xy+3x2−y2)=x2+3x2+5xy−3xy+2y2−y2=4x2+2xy+y2* A−BA - BA−B の計算: A−B=(2y2+5xy+x2)−(−3xy+3x2−y2)=x2−3x2+5xy+3xy+2y2+y2=−2x2+8xy+3y2A - B = (2y^2 + 5xy + x^2) - (-3xy + 3x^2 - y^2) = x^2 - 3x^2 + 5xy + 3xy + 2y^2 + y^2 = -2x^2 + 8xy + 3y^2A−B=(2y2+5xy+x2)−(−3xy+3x2−y2)=x2−3x2+5xy+3xy+2y2+y2=−2x2+8xy+3y2* 2A−3B2A - 3B2A−3B の計算: 2A=2(2y2+5xy+x2)=4y2+10xy+2x22A = 2(2y^2 + 5xy + x^2) = 4y^2 + 10xy + 2x^22A=2(2y2+5xy+x2)=4y2+10xy+2x2 3B=3(−3xy+3x2−y2)=−9xy+9x2−3y23B = 3(-3xy + 3x^2 - y^2) = -9xy + 9x^2 - 3y^23B=3(−3xy+3x2−y2)=−9xy+9x2−3y2 2A−3B=(4y2+10xy+2x2)−(−9xy+9x2−3y2)=2x2−9x2+10xy+9xy+4y2+3y2=−7x2+19xy+7y22A - 3B = (4y^2 + 10xy + 2x^2) - (-9xy + 9x^2 - 3y^2) = 2x^2 - 9x^2 + 10xy + 9xy + 4y^2 + 3y^2 = -7x^2 + 19xy + 7y^22A−3B=(4y2+10xy+2x2)−(−9xy+9x2−3y2)=2x2−9x2+10xy+9xy+4y2+3y2=−7x2+19xy+7y23. 最終的な答え(1)A+B=3x2−2xA + B = 3x^2 - 2xA+B=3x2−2xA−B=−x2+8x+2A - B = -x^2 + 8x + 2A−B=−x2+8x+22A−3B=−4x2+21x+52A - 3B = -4x^2 + 21x + 52A−3B=−4x2+21x+5(2)A+B=x2−3xy−2y2A + B = x^2 - 3xy - 2y^2A+B=x2−3xy−2y2A−B=5x2−5xy+4y2A - B = 5x^2 - 5xy + 4y^2A−B=5x2−5xy+4y22A−3B=12x2−11xy+11y22A - 3B = 12x^2 - 11xy + 11y^22A−3B=12x2−11xy+11y2(3)A+B=4x2+2xy+y2A + B = 4x^2 + 2xy + y^2A+B=4x2+2xy+y2A−B=−2x2+8xy+3y2A - B = -2x^2 + 8xy + 3y^2A−B=−2x2+8xy+3y22A−3B=−7x2+19xy+7y22A - 3B = -7x^2 + 19xy + 7y^22A−3B=−7x2+19xy+7y2