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已知实数a满足a的平方+2a-15=0,求1/(a+1)-(a+2)/(a的平方-1)➗(a+1)(a+2)/(a的平方-2a-1)
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已知实数a满足a的平方+2a-15=0,求1/(a+1)-(a+2)/(a的平方-1)➗(a+1)(a+2)/(a的平方-2a-1)
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答案和解析
a²+2a-15=0
a²+2a+1=16
(a+1)²=16
原式=1/(a+1) -(a+2)/(a+1)(a-1) ×(a-1)²/(a+1)(a+2)
=1/(a+1)-(a-1)/(a+1)²
=(a+1-a+1)/(a+1)²
=2/(a+1)²
=2/16
=1/8
a²+2a+1=16
(a+1)²=16
原式=1/(a+1) -(a+2)/(a+1)(a-1) ×(a-1)²/(a+1)(a+2)
=1/(a+1)-(a-1)/(a+1)²
=(a+1-a+1)/(a+1)²
=2/(a+1)²
=2/16
=1/8
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