已知反比例函数y=k/x和一次函数y=ax+b的图像相交于点A(-2,3)和B(3,m)
求方程k/(x+k)-b/(x+b)=0的根
人气:177 ℃ 时间:2019-10-29 06:15:25
解答
过A
3=k/(-2)
k=-6
则m=-6/3=-2
代入y=ax+b
3=-2a+b
-2=3a+b
相减
5a=-5
a=-1
b=2a+3=1
所以是-6/(x-6)-1/(x+1)=0
-6/(x-6)=1/(x+1)
-6x-6=x-6
x=0
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