已知a²+2a-√3=0,求代数式[(a-2)/(a²+2a)-(a-1)/(a²+4a+4)]÷(a-4)/(a+2)
已知a²+2a-√3=0,求代数式[(a-2)/(a²+2a)-(a-1)/(a²+4a+4)]÷(a-4)/(a+2)
数学人气:136 ℃时间:2019-08-20 13:17:28
优质解答
由[(a-2)/(a(a+2)-(a-1)/(a+2)²]÷(a-4)/(a+2)=[(a-2)(a+2)/a(a+2)²-a(a-1)/a(a+2)²=(a-4)/a(a+2)²×(a+2)/(a-4)=1/a(a+2)∵a²+2a=√3,∴1/a(a+2)=1/(a²+2a)=1/√3=√3/3....
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