高中数学已知实数xy满足x的平方-4x+3+y的平方等于0则x+y+2除以x-1的取值范围是

x²-4x+3+y²=0 令x=x+2→x²+y²=1→y=±√(1-x²) -1≤x≤1(x+y+2)/(x-1)→(x+y+4)/(x+1)（将坐标轴向右平移2个单位，值域不变）令f₁(x)=[x+√(1-x²)+4]/(x+1)-1<x≤1 f₂(x)=[x-√(1-x²)+4]/(x+1)-1<x≤1f₁'(x)=[(x+1)-x(x+1)/√(1-x²)-x-√(1-x²)-4]/(x+1)² =[-x(x+1)/√(1-x²)-√(1-x²)-3]/(x+1)²<0→f₁(x)单调递减→f₁(x)≥f₁(1)=2.5当x→-1时f₁(x)→+∞ f₂'(x)=[(x+1)+x(x+1)/√(1-x²)-x+√(1-x²)-4]/(x+1)² =[x(x+1)/√(1-x²)+√(1-x²)-3]/(x+1)² =[x(x+1)+(1-x²)-3√(1-x²)]/[√(1-x²)(x+1)²] =[x+1-3√(1-x²)]/[√(1-x²)(x+1)²]驻点x=⅘ 左-右+ 为极小值点 极小值f₂(⅘)=2⅓∴(x+y+2)/(x-1)∈[2⅓,+∞)