设f(x)=x-3/x+2 ,求 f(0),f(a+1),f[f(x)]
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f(0)=(0-3)/(0+2)=-3/2
f(a+1)=(a+1-3)/(a+1+2)=(a-2)/(a+3)
f[f(x)]=f(x-3/x+2)
=[(x-3)/(x+2)-3]/[(x-3)/(x+2)+2]
=[(x-3)-3(x+2)]/[(x-3)+2(x+2)]
=(x-3-3x-6)/(x-3+2x+4)
=(-2x-9)/(3x+1)
=-(2x+9)/(3x+1)
f(a+1)=(a+1-3)/(a+1+2)=(a-2)/(a+3)
f[f(x)]=f(x-3/x+2)
=[(x-3)/(x+2)-3]/[(x-3)/(x+2)+2]
=[(x-3)-3(x+2)]/[(x-3)+2(x+2)]
=(x-3-3x-6)/(x-3+2x+4)
=(-2x-9)/(3x+1)
=-(2x+9)/(3x+1)
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