解：f(x)=x³+ax²-(2a+3)x+a²f'(x)=3x²+2ax-(2a+3)f''(x)=6x+2a∵x=1是f(x)极大值点∴f''(1)<0∴a<-3a∈(-∞,-3)希望对你有帮助，望采纳，谢谢~
f'(x)=3x^2+2ax-(2a+3)=3(x-1)[x+(2/3)a+1]-(2/3)a-1>1 a<-3
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