数学 <やや複雑な式の展開>

質問

次の式を展開せよ。
(1)(a+b+3)(a+b-2)

(2)(x+2y+3)(x-2y+3)

(3)(x+y-1)の二乗

(4)(2a-b+3c)の二乗

回答No.1

asuncion

(1)
(a + b + 3)(a + b - 2)
= {(a + b) + 3)}{(a + b) - 2}
= (a + b)^2 + (a + b) - 6
= a^2 + 2ab + b^2 + a + b - 6

(2)
(x + 2y + 3)(x - 2y + 3)
= {x + 3) + 2y}{(x + 3) - 2y}
= (x + 3)^2 - (2y)^2
= x^2 + 6x + 9 - 4y^2

(3)
(x + y - 1)^2
= {(x + y) - 1}^2
= (x + y)^2 - 2(x + y) + 1
= x^2 + 2xy + y^2 - 2x - 2y + 1

(4)
(2a - b + 3c)^2
= {(2a - b) + 3c}^2
= (2a - b)^2 + 6c(2a - b) + (3c)^2
= 4a^2 - 4ab + b^2 + 12ac - 6bc + 9c^2
= 4a^2 + b^2 + 9c^2 - 4ab - 6bc + 12ac