当x=3,y=7时,求分式(xy+2y)-(x-2)分之x²-4的值
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当x=3,y=7时,求分式(xy+2y)-(x-2)分之x²-4的值
当x=3,y=7时,求分式(xy+2y)-(x-2)分之x²-4的值
当x=3,y=7时,求分式(xy+2y)-(x-2)分之x²-4的值
因为你的题目未表达清楚,所以按两种情况考虑
解法1:
当x=3,y=7时:
(xy+2y)-[(x²-4)/(x-2)]
=y(x+2)-(x+2)(x-2)/(x-2)
=y(x+2)-(x+2)
=(y-1)(x+2)
=(7-1)(5+2)
=42
解法2:
当x=3,y=7时:
(x²-4)/[(xy+2y)-(x-2)]
=(x+2)(x-2)/[y(x+2)-(x-2)]
=5/(7*5-1)
=5/34
1/7