$A = 2x^2 + xy - 3z$, $B = -3x^2 + 2xy + z$, $C = x^2 - 3xy + 2z$ であるとき、以下の式を計算する問題です。 (1) $A+B+C$ (2) $A-B+C$ (3) $2A-(B+2C)$ (4) $2(2A+B-C) - (A+4B-C)$

1. 問題の内容

$A = 2x^2 + xy - 3z$, $B = -3x^2 + 2xy + z$, $C = x^2 - 3xy + 2z$ であるとき、以下の式を計算する問題です。

(1) $A+B+C$

(2) $A-B+C$

(3) $2A-(B+2C)$

(4) $2(2A+B-C) - (A+4B-C)$

2. 解き方の手順

(1) $A+B+C$ の計算

$A+B+C = (2x^2 + xy - 3z) + (-3x^2 + 2xy + z) + (x^2 - 3xy + 2z)$

$= (2x^2 - 3x^2 + x^2) + (xy + 2xy - 3xy) + (-3z + z + 2z)$

$= (2-3+1)x^2 + (1+2-3)xy + (-3+1+2)z$

$= 0x^2 + 0xy + 0z$

$= 0$

(2) $A-B+C$ の計算

$A-B+C = (2x^2 + xy - 3z) - (-3x^2 + 2xy + z) + (x^2 - 3xy + 2z)$

$= (2x^2 + xy - 3z) + (3x^2 - 2xy - z) + (x^2 - 3xy + 2z)$

$= (2x^2 + 3x^2 + x^2) + (xy - 2xy - 3xy) + (-3z - z + 2z)$

$= (2+3+1)x^2 + (1-2-3)xy + (-3-1+2)z$

$= 6x^2 - 4xy - 2z$

(3) $2A-(B+2C)$ の計算

$2A = 2(2x^2 + xy - 3z) = 4x^2 + 2xy - 6z$

$2C = 2(x^2 - 3xy + 2z) = 2x^2 - 6xy + 4z$

$B+2C = (-3x^2 + 2xy + z) + (2x^2 - 6xy + 4z) = (-3+2)x^2 + (2-6)xy + (1+4)z$

$B+2C = -x^2 - 4xy + 5z$

$2A - (B+2C) = (4x^2 + 2xy - 6z) - (-x^2 - 4xy + 5z)$

$= (4x^2 + 2xy - 6z) + (x^2 + 4xy - 5z)$

$= (4+1)x^2 + (2+4)xy + (-6-5)z$

$= 5x^2 + 6xy - 11z$

(4) $2(2A+B-C) - (A+4B-C)$ の計算

$2A = 4x^2 + 2xy - 6z$

$2A + B - C = (4x^2 + 2xy - 6z) + (-3x^2 + 2xy + z) - (x^2 - 3xy + 2z)$

$= (4-3-1)x^2 + (2+2+3)xy + (-6+1-2)z$

$= 0x^2 + 7xy - 7z = 7xy - 7z$

$2(2A+B-C) = 2(7xy - 7z) = 14xy - 14z$

$4B = 4(-3x^2 + 2xy + z) = -12x^2 + 8xy + 4z$

$A+4B-C = (2x^2 + xy - 3z) + (-12x^2 + 8xy + 4z) - (x^2 - 3xy + 2z)$

$= (2-12-1)x^2 + (1+8+3)xy + (-3+4-2)z$

$= -11x^2 + 12xy - z$

$2(2A+B-C) - (A+4B-C) = (14xy - 14z) - (-11x^2 + 12xy - z)$

$= (14xy - 14z) + (11x^2 - 12xy + z)$

$= 11x^2 + (14-12)xy + (-14+1)z$

$= 11x^2 + 2xy - 13z$

3. 最終的な答え

(1) 0

(2) $6x^2 - 4xy - 2z$

(3) $5x^2 + 6xy - 11z$

(4) $11x^2 + 2xy - 13z$