(5)一般型の展開 2(解答) 学習日    月    日(   )
 乗法公式 3(一般型の展開)     (x+2y)(x+3y)
  =x2+(2+3)xy+2・3y2
  =x2+5xy+6y2 
(xay)(xby)=x2+(ab)xyaby2
           和    積
 
乗法公式を使って,次の式を展開しなさい。
1  (x+2y)(x+4y) = x2+6xy+8y2
 
 (x+5y)(x+7y) = x2+12xy+35y2
 
 (x+4y)(xy) = x2+5xy+4y2
 
 (x+8y)(x+3y) = x2+11xy+24y2
 
 (a+3b)(a+2b) = a2+5ab+6b2
 
 (ab)(a+5b) = a2+6ab+5b2
 
 (x-2y)(x-6y) = x2-8xy+12y2
  
 (x-5y)(x-4y) = x2-9xy+20y2
  
 (a-7b)(a-9b) = a2-16ab+63b2
 
10  (ab)(a-2b) = a2-3ab+2b2
 
11  (x+4y)(x-2y) = x2+2xy-8y2
  
12  (xy)(x+2y) = x2xy-2y2
  
13  (xy)(x-3y) = x2-2xy-3y2
  
14  (x-6y)(x-8y) = x2-14xy+48y2
  
15  (a-7b)(a+4b) = a2-3ab-28b2
 
16  (a+20b)(a+30b) = a2+50ab+600b2
 
17  (2x+3)(2x+5) = (2x)2+8(2x)+15
  = 4x2+16x+15
  
18  (3x+1)(3x-2) = (3x)2-(3x)-2
  = 9x2-3x-2
  
19  (5a-1)(5a-2) = (5a)2-3(5a)+2
  = 25a2-15a+2
  
20  (2y-7)(2y+4) = (2y)2-3(2y)-28
  = 4y2-6y-28
  

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