若 √﹙a²-3a+1﹚+b²+2b+1=0,则=a²+﹙1/a²﹚-|b|=____.
问题描述:
若 √﹙a²-3a+1﹚+b²+2b+1=0,则=a²+﹙1/a²﹚-|b|=____.
根号下管到a²-3a+1
﹙1/a²﹚的意思是a²分之1
|b|是b的绝对值
____请输入得数
答
√﹙a²-3a+1﹚+b²+2b+1=0 √﹙a²-3a+1﹚+﹙b+1﹚²=0∴a²-3a+1=0,b+1=0a-3+1/a=0,b=-1a+1/a=3a²+1/a²=﹙a+1/a﹚²-2=9-2=7a²+﹙1/a²﹚-|b...