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已知x+y=0,x+3y=1,求3x^2+12xy+13y^2 的值
人气:465 ℃ 时间:2020-05-14 18:44:11
解答
3x^2+12xy+13y^2
=x^2+2xy+y^2 + 2x^2+10xy+12y^2
=(x+y)^2 + 2(x+2y)(x+3y)
=0+2(x+2y)*1
=0+2(x+y + y)
=2y
=(x+3y)-(x+y)
=1-0
=1
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