$A = 2x^2 + xy - 3z$, $B = -3x^2 + 2xy + z$, $C = x^2 - 3xy + 2z$ が与えられたとき、$2A - (B + 2C)$ を計算せよ。代数学多項式式の計算代数2025/5/291. 問題の内容A=2x2+xy−3zA = 2x^2 + xy - 3zA=2x2+xy−3z, B=−3x2+2xy+zB = -3x^2 + 2xy + zB=−3x2+2xy+z, C=x2−3xy+2zC = x^2 - 3xy + 2zC=x2−3xy+2z が与えられたとき、2A−(B+2C)2A - (B + 2C)2A−(B+2C) を計算せよ。2. 解き方の手順まず、2A2A2A、2C2C2C、B+2CB+2CB+2C を計算し、その後 2A−(B+2C)2A-(B+2C)2A−(B+2C) を計算する。2A=2(2x2+xy−3z)=4x2+2xy−6z2A = 2(2x^2 + xy - 3z) = 4x^2 + 2xy - 6z2A=2(2x2+xy−3z)=4x2+2xy−6z2C=2(x2−3xy+2z)=2x2−6xy+4z2C = 2(x^2 - 3xy + 2z) = 2x^2 - 6xy + 4z2C=2(x2−3xy+2z)=2x2−6xy+4zB+2C=(−3x2+2xy+z)+(2x2−6xy+4z)=(−3+2)x2+(2−6)xy+(1+4)z=−x2−4xy+5zB + 2C = (-3x^2 + 2xy + z) + (2x^2 - 6xy + 4z) = (-3+2)x^2 + (2-6)xy + (1+4)z = -x^2 - 4xy + 5zB+2C=(−3x2+2xy+z)+(2x2−6xy+4z)=(−3+2)x2+(2−6)xy+(1+4)z=−x2−4xy+5z2A−(B+2C)=(4x2+2xy−6z)−(−x2−4xy+5z)=(4−(−1))x2+(2−(−4))xy+(−6−5)z=(4+1)x2+(2+4)xy+(−11)z=5x2+6xy−11z2A - (B + 2C) = (4x^2 + 2xy - 6z) - (-x^2 - 4xy + 5z) = (4-(-1))x^2 + (2-(-4))xy + (-6-5)z = (4+1)x^2 + (2+4)xy + (-11)z = 5x^2 + 6xy - 11z2A−(B+2C)=(4x2+2xy−6z)−(−x2−4xy+5z)=(4−(−1))x2+(2−(−4))xy+(−6−5)z=(4+1)x2+(2+4)xy+(−11)z=5x2+6xy−11z3. 最終的な答え5x2+6xy−11z5x^2 + 6xy - 11z5x2+6xy−11z