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已知m²+n²=5,x²+y²=401,求多项式（mx=ny)²+（nx-my)²的值
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已知m²+n²=5,x²+y²=401,求多项式（mx=ny)²+（nx-my)²的值
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答案和解析
（mx+ny)²+（nx-my)²
=m²x²+n²y²+n²x²+m²y²
=(m²+n²)(x²+y²)
=5×401
=2005
=m²x²+n²y²+n²x²+m²y²
=(m²+n²)(x²+y²)
=5×401
=2005
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