搜搜考试网SOS 谁能帮我解数学题,
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SOS 谁能帮我解数学题,
已知(x-3)的平方加上3y+1的绝对值等于0.求3xX的平方y-[2xyY的平方-2(xy-3\2)+xy)+3xyY的平方的值.
已知(x-3)的平方加上3y+1的绝对值等于0.求3xX的平方y-[2xyY的平方-2(xy-3\2)+xy)+3xyY的平方的值.
(x-3)^2+|3y+1|=0
(x-3)^2=0,x=3
|3y+1|=0.y=-1/3
3x^2y-[2xy^2-2(xy-3/2)+xy]+3xy^2
=3x^2y-[2xy^2-2xy+3/2+xy]+3xy^2
=3x^2y-[2xy^2-xy+3/2]+3xy^2
=3x^2y-2xy^2+xy-3/2+3xy^2
=x^2y+xy+3xy^2-3/2
=xy(x+1+3y)-3/2
=-3*1/3(3+1-3*1/3)-3/2
=-3-3/2
=-9/2
(x-3)^2=0,x=3
|3y+1|=0.y=-1/3
3x^2y-[2xy^2-2(xy-3/2)+xy]+3xy^2
=3x^2y-[2xy^2-2xy+3/2+xy]+3xy^2
=3x^2y-[2xy^2-xy+3/2]+3xy^2
=3x^2y-2xy^2+xy-3/2+3xy^2
=x^2y+xy+3xy^2-3/2
=xy(x+1+3y)-3/2
=-3*1/3(3+1-3*1/3)-3/2
=-3-3/2
=-9/2