用二次函数换元法解y=2x+√2-√x用分离常数法解y=x²-x/x²-x+2

Question

用二次函数换元法解 y=2x+√2-√x 用分离常数法解 y=x²-x/x²-x+2

Answer 1

1.是求y=2x+√2-√x 的值域吧?
令t=√x,则t≥0,
y=2t² -t +√2= 2(t-1/4)²+√2 -1/8
∵t≥0,∴ y∈[√2 -1/8,+∞)
2.是这样的吧?y=(x²-x)/(x²-x+2)
y=(x²-x)/(x²-x+2)=[(x²-x+2) -2]/(x²-x+2)=1 - 2/(x²-x+2)
而x²-x+2=(x-1/2)²+ 7/4≥7/4
∴0