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已知(a+2)(根号10+1)=1,求1+1/a-1-(1+2a)/(a²+a-2)的值
人气:134 ℃ 时间:2020-04-03 23:42:18
解答
∵a+2=1/(√10+1)=(√10-1)/9
∴原式=1+1/(a-1)-(1+2a)/(a+2)(a-1)
=(a²+a-2+a+2-1-2a)/(a+2)(a-1)
=(a²-1)/(a+2)(a-1)
=(a-1)/(a+2)
=1-3/(a+2)
=1-(√10-1)/9
=(9-√10+1)/9
=(10-√10)/9
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