問題
$A = 2x^2 + xy – 3z,\ B = -3x^2 + 2xy + z,\ C = x^2 – 3xy + 2z$ であるとき, 次の式を計算せよ.
解説
- (1) $A + B + C$
- $2x^2 + xy – 3z+(-3x^2 + 2xy + z)+x^2 – 3xy + 2z$
$=$$0$
- $2x^2 + xy – 3z+(-3x^2 + 2xy + z)+x^2 – 3xy + 2z$
- (2) $A – B + C$
- $2x^2 + xy – 3z-(-3x^2 + 2xy + z)+x^2 – 3xy + 2z$
$=$$6x^2-4xy-2z$
- $2x^2 + xy – 3z-(-3x^2 + 2xy + z)+x^2 – 3xy + 2z$
- (3) $2A – (B + 2C)$
- $2(2x^2 + xy – 3z)-\{-3x^2 + 2xy + z+2(x^2 – 3xy + 2z)\}$
$=$$5x^2+6xy-11z$
- $2(2x^2 + xy – 3z)-\{-3x^2 + 2xy + z+2(x^2 – 3xy + 2z)\}$
- (4) $2(2A + B – C) – (A + 4B – C)$
- $=3A-2B-C$より,
$3(2x^2 + xy – 3z)-2(-3x^2 + 2xy + z)-(x^2 – 3xy + 2z)$
$=$$11x^2+2xy-13z$
- $=3A-2B-C$より,