∵x²+1>0
y²-2y+3=y²-2y+1+2=(y-1)²+2>0
又∵|(3a-b-4)(x^2+1)|+|(4a+b-3)(y^2-2y+3)|=0
∴3a-b-4=0
4a+b-3=0
∴a=1
b=-1
∴|2a|-|3b|
=2-3
=-1
若|(3a-b-4)(x^2+1)|+|(4a+b-3)(y^2-2y+3)|=0,求|2a|-3|b|.
若|(3a-b-4)(x^2+1)|+|(4a+b-3)(y^2-2y+3)|=0,求|2a|-3|b|.
數(shù)學(xué)人氣：530 ℃時(shí)間：2020-03-29 16:46:55
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