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(1,-2), (-4,-2)及下列四条曲线: ①4x+2y=3 ②x2+y2=3 ③x2+2y2=3 ④x2-2y2=3 其中曲线上存在点P.,使|PA|=|PB|的曲线有( ) A.①③B.②④ ①②③ ②③④
y3=2(y-xy′) 2xy′=2y 2xy′=-y3 2xy=2y+y3
y=﹣4x+3 y=﹣4x﹣3 y=4x+3 y=4x﹣3
y-3x+1=0 x-y+3=0 2x-y-3=0 2x-y+3=0
y-3x+1=0 y-3x=0 y-3x-1=0 3y-3x-1=0
y+3x+6=0 y-3x-1=0 y-3x-8=0 y+3x+1=0
y-3x+1=0 y-3x=0 y-3x-1=0 3y-3x-1=0
x+y-2=0 3x+y-2=0 3x-y-2=0 x-y+2=0
y+3x+b=0 y-3x-1=0 y-3x-8=0 y+3x+1=0
y=x3 y=x3+c y=x3+2 y=x3+4
y=x3-2B y=2x3-5 y=x2-2D y=2x2-5
y-3x+1=0 x-y+3=0 2x-y-3=0 2x-y+3=0
y-3x+1=0 x-y+3=0 2x-y-3=0 2x-y+3=0
y=3x-1 y=-3x+5 y=3x+5 y=2x