f(x)=(m-1)x²+2mx+3是在[-3a+1,2]是偶函数定义域关于原点对称则-3a+1+2=0解得a=1定义域[-2,2]f(-x)=(m-1)x²-2mx+3f(x)=f(-x)所以(m-1)x²+2mx+3=(m-1)x²-2mx+3即2m=-2m所以解得m=0...