解： x²+y²=4 => 1+y²/x²=4/x² => 1/x²=1/4+y²/4x² ∴1/x²-y/x =1/4+y²/4x²-y/x =y²/4x²-y/x+1/4 =1/4·(y/x-2)²-3/4 1+y²/x²=4/x²≥1 => 2≥x≥-2 => 2≥y≥-2 当y/x=2时1/x²-y/x取最小值-3/4. 1/x²-y/x的取值范围是[-3/4,∞). 或者利用圆的参数方程 设x=2cθ, y=2sinθ θ∈[0,2π] 则1/x²-y/x =(sin²θ+cos²θ)/4cos²θ-sinθ/cosθ =sin²θ/4cos²θ+1/4-sinθ/cosθ =1/4·(sinθ/cosθ-2)²-3/4 =1/4·(tanθ-2)²-3/4 ∵tanθ∈(-∞,∞) ∴当tanθ=2时取最小值-3/4 ∴1/x²-y/x∈[-3/4,∞)