解方程组(x+2)(y-1)=(x-2)(y+2)和(x-12)(y-14)=(x+12)(y-2)
问题描述:
解方程组(x+2)(y-1)=(x-2)(y+2)和(x-12)(y-14)=(x+12)(y-2)
答
(x+2)(y-1)=(x-2)(y+2)
(x-12)(y-14)=(x+12)(y-2)
xy+2y-x-2=xy-2y+2x-4
xy-12y-14x+168=xy+12y-2x-24
3x-4y-2=0
12x+24y-192=0
3x-4y-2=0
x+2y-16=0
{x=34/5
{y23/5不好意思我题目打错了(x+2)(y-1)=(x-2)(y+3)和(x-12)(y-14)=(x+12)(y-2)是这个再帮下忙(x+2)(y-1)=(x-2)(y+3)和(x-12)(y-14)=(x+12)(y-2) (x+2)(y-1)=(x-2)(y+3)(x-12)(y-14)=(x+12)(y-2)xy+2y-x-2=xy-2y+3x-6xy-12y-14x+168=xy+12y-2x-244x-4y-4=012x+24y-192=0x-y-1=0x+2y-16=0下式减上式得:3y=15y=5x=6所以原式的解为:{x=6{y=5答案已验证确认正确!