
解:∵abc不等于0 且a+b+c不等于0c/(a+b)=a/(b+c)=b/(a+c)由等比性质,得(a+b+c)/(2a+2b+2c)=k∴k=(a+b+c)/(2a+2b+2c) =(a+b+c)/[2(a+b+c)] =½∴k-1=½-1=-½则反比例函数y=(k-1)/x的图像经过【第二、四】象限.

解:∵abc不等于0 且a+b+c不等于0c/(a+b)=a/(b+c)=b/(a+c)由等比性质,得(a+b+c)/(2a+2b+2c)=k∴k=(a+b+c)/(2a+2b+2c) =(a+b+c)/[2(a+b+c)] =½∴k-1=½-1=-½则反比例函数y=(k-1)/x的图像经过【第二、四】象限.