求出值域:1.y=(3+x)/(4-x) 2.y=5/2x^2-4x+3
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求出值域:1.y=(3+x)/(4-x) 2.y=5/2x^2-4x+3
求出值域:1.y=(3+x)/(4-x) 2.y=5/2x^2-4x+3
求出值域:1.y=(3+x)/(4-x) 2.y=5/2x^2-4x+3
y=(3+x)/(4-x)=-(x+3)/(x-4)
=-(x-4+7)/(x-4)
=-[(x-4)/(x-4)+7/(x-4)]
=-[1+7/(x-4)]
=-1-7/(x-4)
因为7/(x-4)≠0
所以y≠-1-0=-1
所以值域(-∞,-1)∪(-1,+∞)
2x^2-4x+3=2(x-1)^2+1>=1
所以0