x^2+y^2+4x-4y-10=0,C2:x^2+y^2-4x+4y+6=0,证明这两圆相切,并求过切点的切线方程
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如果A={(x,y)|x,y+2>0},B={(x,y)|2x+3y-6>0},C={(x,y)|x-4
x*x+2x+y*y-4y+5=0 x= y=
{4x-3y-10=0 3x-2y=0,{3x-4(x-y)=2 2x-3y=1,{2(x+y)-(x-y)=3 (x+y)-2(x-y)=1
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(x-y) (x-2y)+(x+2y) (x-3y)-2(x-3y) (x-4y)=
{2x+y=10 {x-y+z=4 {3x-y-z=0
x²+4y²-2x+12y+10=0求X,Y值
若x.y满足条件2x+y-12=0,x-4y+10
(x-4y)(2x+y)=
圆C:x^2+y^2-2x+4y-15=0 上有一点(x,y) 求2x+y的最大值和最小值圆C:x^2+y^2-2x+4y-15=0上有一点(x,y)求2x+y的最大值和最小值答案是10,-10 感恩
(4x-2y-z)-{5x[8y-2y-(x+y)]-x+(3y-10z)]=? kuaihuajian
已知x^2-4x+y^2-10y+29=0,求x^2y^2+2x^3y^2+x^4y^2的值
若x^2-4x+y^2-10y+29=0,求x^2y^2+2x^3y^2+x^4y^2的值
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