求y=2x/x^2-x+1的值域

函数y=2x/(x²-x+1)中,∵x²-x+1=(x-1/2)²+(3/4)≥3/4>0,∴x²-x+1恒不为0,函数f(x)= 2x/(x²-x+1)的定义域为R.函数y的解析式y=2x/(x²-x+1)可化为(x²-x+1)y=2xyx²-(y+2)x+y=0（...