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为简单起见,不妨设A=1,
x^2+2Bxy+Cy^2+2Dx+2Ey+F=(x+my+n)(x+py+q)
=x^2+(m+p)xy+mpy^2+(n+q)x+(mq+np)y+nq,
比较系数得
m+p=2B,mp=C,
∴m,p=B土√(B^2-C),
n+q=2D,nq=F,
∴n,q=D土√(D^2-F)
代入mq+np=2E,得
[B+√(B^2-C)][D+√(D^2-F)]+[B-√(B^2-C)][D-√(D^2-F)]=2E,或
[B+√(B^2-C)][D-√(D^2-F)]+[B-√(B^2-C)][D+√(D^2-F)]=2E,为所求.