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令x+1=t,则x=a-1
f(t)=(t-1)^2-3(t-1)+2
=t^2-5t+6
f(x)=x^2-5x+6
f(2)=0
f(a)=a^2-5a+6
f(x-1)=(x-1)^2-5(x-1)+6
=x^2-7x+12
f(x)=2x-3
{x∈N|1≤x≤5}
f(x)∈{-1,1,3,5,7}
y=-x^2+9
x∈[-2,3]
{y|0≤y≤9}
y=x/(x-3)=1+[3/(x-3)]
x∈[4,7]
{y|7/4≤y≤4}
f(t)=(t-1)^2-3(t-1)+2
=t^2-5t+6
f(x)=x^2-5x+6
f(2)=0
f(a)=a^2-5a+6
f(x-1)=(x-1)^2-5(x-1)+6
=x^2-7x+12
f(x)=2x-3
{x∈N|1≤x≤5}
f(x)∈{-1,1,3,5,7}
y=-x^2+9
x∈[-2,3]
{y|0≤y≤9}
y=x/(x-3)=1+[3/(x-3)]
x∈[4,7]
{y|7/4≤y≤4}
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