设A=2x^2-3xy+y^2-x+2y,B=4x^2-6xy+2y^2-3x-y 若/x-2a/+(y+3)^2=0且
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设A=2x^2-3xy+y^2-x+2y,B=4x^2-6xy+2y^2-3x-y 若/x-2a/+(y+3)^2=0且B-2A=a,求A的值?
|x+2a|+(y+3)^2=0
x=-2a,y=-3
B-2A=a
a=4x^2-6xy+2y^2-3x-y-2(2x^2-3xy+y^2-x+2y)
=4x^2-6xy+2y^2-3x-y-4x^2+6xy-2y^2+2x-4y
=-x-5y
|x-2a|
=|x-2(-x-5y)|
=|x+2x+10y|
=|3x+10y|
=|3x-30|
=0
解得x=10
A=2x^2-3xy+y^2-x+2y
=(x-y)(2x-y)-x+2y
=13×23-10-6
=283
x=-2a,y=-3
B-2A=a
a=4x^2-6xy+2y^2-3x-y-2(2x^2-3xy+y^2-x+2y)
=4x^2-6xy+2y^2-3x-y-4x^2+6xy-2y^2+2x-4y
=-x-5y
|x-2a|
=|x-2(-x-5y)|
=|x+2x+10y|
=|3x+10y|
=|3x-30|
=0
解得x=10
A=2x^2-3xy+y^2-x+2y
=(x-y)(2x-y)-x+2y
=13×23-10-6
=283
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