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若X是整数,求证:(X+2)(X+4)(X+6)(X+8)+16是一个完全平方数
人气:162 ℃ 时间:2020-05-13 21:52:50
解答
(X+2)(X+4)(X+6)(X+8)+16=(x+5-3)(x+5-1)(x+5+1)(x+5+3) + 16= {(x+5)^2-3^2}{(x+5)^2-1^2} + 16= (x+5)^2-10(x+5)+9+16= (x+5)^2-10(x+5)+25= (x+5-5)^2= x^2,得证2x2+26xy-60y2(X+2)(X+4)(X+6)(X+8)+16=(x+5-3)(x+5-1)(x+5+1)(x+5+3) + 16= {(x+5)^2-3^2}{(x+5)^2-1^2} + 16= (x+5)^4-10(x+5)^2+9+16= (x+5)^4-10(x+5)^2+25= {(x+5)^2-5}^2得证。2x2+26xy-60y2= 2{x^2+13x-30y^2}= 2(x+15y)(x-2y)x2(x+y)+x(y2+1)(x+y)+y2x2(x+y)+x(y2+1)(x+y)+y2?这个想干什么?你确定没有写错吗先采纳了再说~a2+4b2-4ab-4b+2a+1=a^2-4ab+4b^2+2a-4b+1= (a-2b)^2 + 2(a-2b) + 1= (a-2b+1)^2
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