x^2+y^2+4x+6y+13=0,且x、y为实数
x^2+4x+4+y^2+6y+9=0
(x+2)^2+(y+3)^2=0
x+2=0 y+3=0
x=-2 y=-3
xy^2
=(-2)(-3)^2
=-18
巳知x^2+y^2+4x+6y+13=0,且x、y为实数.则xy^2=?
巳知x^2+y^2+4x+6y+13=0,且x、y为实数.则xy^2=?
数学人气:729 ℃时间:2020-03-25 06:59:28
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