A、B兩點(diǎn)關(guān)于原點(diǎn)對(duì)稱,可得：
-4+a^2=0 解得：a=2或-2
3a-b+2b-1=0 解得：b=1-3a 即：b=7 或 b=-5
又因點(diǎn)A在第三象限,所以點(diǎn)B在第一象限,可得：
2b-1>0 解得：b>1/2
綜上可得：a=-2,b=7
若A(3a-b.-4)與點(diǎn)B(2b-1,a的平方)關(guān)于原點(diǎn)對(duì)稱,且點(diǎn)A在第三象限,a= b=
若A(3a-b.-4)與點(diǎn)B(2b-1,a的平方)關(guān)于原點(diǎn)對(duì)稱,且點(diǎn)A在第三象限,a= b=
數(shù)學(xué)人氣：887 ℃時(shí)間：2019-12-06 20:22:10
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