y=(x²-2x+5)/(x-1)分式有意义,x-1≠0 x≠1,即x可取除1以外任意实数.整理,得x²-(y+2)x+y+5=0令x=1,得1-(y+2)+y+5=0 4=0,等式恒不成立,即方程不会产生根为x=1方程有解,判别式△≥0[-(y+2)]²-4(y+5)≥0y...