已知p²-2p-5=0,5q²+2q-1=0,其中p、q为实数,求p²+1/(q²)
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![已知p²-2p-5=0,5q²+2q-1=0,其中p、q为实数,求p²+1/(q²)](/uploads/image/z/1104644-20-4.jpg?t=%E5%B7%B2%E7%9F%A5p%26%23178%3B-2p-5%3D0%2C5q%26%23178%3B%2B2q-1%3D0%2C%E5%85%B6%E4%B8%ADp%E3%80%81q%E4%B8%BA%E5%AE%9E%E6%95%B0%2C%E6%B1%82p%26%23178%3B%2B1%2F%28q%26%23178%3B%29)
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已知p²-2p-5=0,5q²+2q-1=0,其中p、q为实数,求p²+1/(q²)
已知p²-2p-5=0,5q²+2q-1=0,其中p、q为实数,求p²+1/(q²)
已知p²-2p-5=0,5q²+2q-1=0,其中p、q为实数,求p²+1/(q²)
5q平方 2q-1=0,1/q^2-2/q-5=0
所以,p和1/q是方程 x^2-2x-5=0的两个实数根,
p+1/q=2,p^2+1/q^2+2=4,p^2+1/q^2=2