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∵f(x)=(m-1)x2+mx+1是偶函數,∴m=0,∴f(x)=-x2+1則f(x)在區間[-2,1]上的最大值與最小值分別為-3和1則f(x)在區間[-2,1]上的最大值與最小值的和等於-2故答案為:-2
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