初中一找规律的题,在线等,速度,x与y同号,且是方程组|x|+|y|=7,2|x|-3|y|=-1的解试求代数式,1(x+5)(y+5)+1(x+6)(y+6)+.+1(x+28)(y+28)的值要过程,谢谢了

Question

初中一找规律的题,在线等,速度,
x与y同号,且是方程组|x|+|y|=7,2|x|-3|y|=-1的解
试求代数式,1\(x+5)(y+5)+1\(x+6)(y+6)+.+1\(x+28)(y+28)的值
要过程,谢谢了

Answer 1

|x|+|y|=7
2|x|+2|y|=14.1
2|x|-3|y|=-1.2
1式-2式得
5|y|=15
|y|=3
|x|+|y|=7
|x|+3=7
|x|=4
因为x与y同号
所以x=4,y=3或x=-4,y=-3
当x=4,y=3时
1/(x+5)(y+5)+1/(x+6)(y+6)+.+1/(x+28)(y+28)
=1/(8*9)+1/(9*10)+.+1/(31*32)
=1/8-1/9+1/9-1/10+.+1/31-1/32
=1/8-1/32
=4/32-1/32
=3/32
当x=-4,y=-3时
1/(x+5)(y+5)+1/(x+6)(y+6)+.+1/(x+28)(y+28)
=1/(1*2)+1/(2*3)+.+1/(24*25)
=1-1/2+1/2-1/3+.+1/24-1/25
=1-1/25
=24/25