Question

已知x≠-1,则代数式x^2/(x+1)的取值范围

快，要详细过程，急

Answer 1

x²=(x+1)²-2(x+1)+1.∴代数式y=x²/(x+1)=(x+1)+[1/(x+1)-2.===>y+2=(x+1)+[1/(x+1)].∵x≠-1.∴x+1≠0.(1)当x+1＞0时，由基本不等式得y+2≥2.===>y≥0.（2）当x+1＜0时，-(y+2)=[-(x+1)]+[1/(-x-1)]≥2.===>y≤-4,∴值域为(-∞,-4]∪[0,+∞).【用判别式法也很好】y=x²/(x+1).===>x²-yx-y=0.⊿=y²+4y≥0.===>y(y+4)≥0.===>y≤-4,y≥0.