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由f(x)=ax^2+bx得f(-1)=a-b①f(1)=a+b②f(-2)=4a-2b由①+②得a=[f(1)+f(-1)],由②-①得b=[f(1)-f(-1)]從而f(-2)=2[f(1)+f(-1)]-[f(1)-f(-1)]=3f(-1)+f(1)又1≤f(一1)≤2,2…
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