(3)式の展開(解答) | 学習日 月 日( ) | ||||
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分配法則を使って,次の式を展開しなさい。 | |||
1 | (a+2)(b+3) = ab+3a+2b+6 |
2 | (a+5)(b+4) = ab+4a+5b+20 |
3 | (a+4)(b−6) = ab−6a+4b−24 |
4 | (a−9)(b+2) = ab+2a−9b−18 |
5 | (x+1)(y+6) = xy+6x+y+6 |
6 | (x+8)(y+2) = xy+2x+8y+16 |
7 | (x−7)(y−3) = xy−3x−7y+21 |
8 | (x+3)(y−4) = xy−4x+3y−12 |
9 | (a+b)(c−d) = ac−ad+bc−bd |
10 | (a−b)(x−y) = ax−ay−bx+by |
11 | (2x+1)(y+3) = 2xy+6x+y+3 |
12 | (x−2)(5y−4) = 5xy−4x−10y+8 |
13 | (3a+5)(b+2) = 3ab+6a+5b+10 |
14 | (a+1)(4a−1) = 4a2−a+4a−1 = 4a2+3a−1 |
15 | (x+8)(x+3) = x2+3x+8x+24 = x2+11x+24 |
16 | (a−2)(a+3) = a2+3a−2a−6 = a2+a−6 |
17 | (x−7)(x+7) = x2+7x−7x−49 = x2−49 |
18 | (3x+1)(3x+4) =9x2+12x+3x+4 = 9x2+15x+4 |
19 | (3a−2b)(a+4b)=3a2+12ab−2ab−8b2 = 3a2+10ab−8b2 |
20 | (x+y)(x−y+1)=x2−xy+x+xy−y2+y = x2+x−y2+y |
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