f(x)=(x²+2x+1)/(x-1)
=(x²-2x+1+4x-4+4)/(x-1)
=[(x-1)²+4(x-1)+4]/(x-1)
=(x-1)+4+4/(x-1)
因为x＞1,所以x-1＞0
故(x-1)+4/(x-1)≥2√[(x-1)×4/(x-1)]=4,此时x=3
故f(x)≥4+4=8
所以值域为[8,+∞)
求f(x)=x2+2x+1/x-1(x>1)的值域
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f(x)=(x²+2x+1)/(x-1)
=(x²-2x+1+4x-4+4)/(x-1)
=[(x-1)²+4(x-1)+4]/(x-1)
=(x-1)+4+4/(x-1)
因为x＞1,所以x-1＞0
故(x-1)+4/(x-1)≥2√[(x-1)×4/(x-1)]=4,此时x=3
故f(x)≥4+4=8
所以值域为[8,+∞)