有关因式分解的题求[(x^2+y^2)(a^2+b^2)+4abxy]^2-4[xy(a^2+b^2)+ab(x^2+y
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有关因式分解的题
求[(x^2+y^2)(a^2+b^2)+4abxy]^2-4[xy(a^2+b^2)+ab(x^2+y^2)]^2因式分解的结果
求[(x^2+y^2)(a^2+b^2)+4abxy]^2-4[xy(a^2+b^2)+ab(x^2+y^2)]^2因式分解的结果
[(x^2+y^2)(a^2+b^2)+4abxy]^2-4[xy(a^2+b^2)+ab(x^2+y^2)]^2
设x²+y²=m,a²+b²=n,xy=p,ab=q,则原式化为
(nm+4pq)²-4(pn+qm)²
=(nm+4pq+2pn+2qm)(nm+4pq-2pn-2qm)
=[n(m+2p)+2q(m+2p)][n(m-2p)-2q(m-2p)]
=(m+2p)(n+2q)(m-2p)(n-2q)
=(x²+y²+2xy)(a²+b²+2ab)(x²+y²-2xy)(a²+b²-2ab)
=(x+y)²(a+b)²(x-y)²(a-b)²
说明:换元法
设x²+y²=m,a²+b²=n,xy=p,ab=q,则原式化为
(nm+4pq)²-4(pn+qm)²
=(nm+4pq+2pn+2qm)(nm+4pq-2pn-2qm)
=[n(m+2p)+2q(m+2p)][n(m-2p)-2q(m-2p)]
=(m+2p)(n+2q)(m-2p)(n-2q)
=(x²+y²+2xy)(a²+b²+2ab)(x²+y²-2xy)(a²+b²-2ab)
=(x+y)²(a+b)²(x-y)²(a-b)²
说明:换元法
有关因式分解的题求[(x^2+y^2)(a^2+b^2)+4abxy]^2-4[xy(a^2+b^2)+ab(x^2+y
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