已知a+1/a=根号10,求(1)a²+1/a² (2)a²/a^4+a^2+1急
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已知a+1/a=根号10,求(1)a²+1/a² (2)a²/a^4+a^2+1急
已知a+1/a=根号10,求(1)a²+1/a² (2)a²/a^4+a^2+1
急
已知a+1/a=根号10,求(1)a²+1/a² (2)a²/a^4+a^2+1急
解
a+1/a=√10
两边平方
(a+1/a)²=10
即
a²+1/a²+2=10
∴a²+1/a²=8
是这个吗?
a²/(a^4+a^2+1)
=1/(a²+1+1/a²)
=1/(8+1)
=1/9
还是这个
(a²/a^4)+a²+1
=1/a²+a²+1
=9