Question

y=(2x2一x十2)/x2十x+1的值域

Answer 1

YX^2+YX+Y=2X^2-X+2

(Y-2)X^2+(Y+2)X+(Y-2)=0,

当Y=2时,X=0,

当Y≠2时,

Δ=(Y+2)^2-4(Y-2)

=(Y+2+2Y-4)(Y+2-2Y+4)

=(3Y-2)(-Y+6)≥0,

得：2/3≤Y≤6,

∴值域：2/3≤Y≤6.