若|p+2|与q2-8q+16互为相反数,分解因式(x2+y2)-(pxy+q)=______.
人气:303 ℃ 时间:2020-05-12 10:01:32
解答
依题意得|p+2|+(q2-8q+16)=0,即|p+2|+(q-4)2=0,
∴p+2=0,q-4=0,
解得p=-2,q=4,
∴(x2+y2)-(pxy+q),
=(x2+y2)-(-2xy+4),
=x2+y2+2xy-4,
=(x2+2xy+y2)-4,
=(x+y)2-22,
=(x+y+2)(x+y-2).
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