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∵a、b是關於x的方程x2+(m-2)x+1=0的兩根,∴a2+(m-2)a+1=0,b2+(m-2)b+1=0,ab=1,∴1+ma+a2=2a,1+mb+b2=2b,∴(1+ma+a2)(1+mb+b2)=2a•2b=4ab=4.故答案為:4.
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