已知圆(x-3)2 (y-2)2 25,过点(1,0)做两条互相垂直的弦

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/14 23:22:46
已知圆(x-3)2 (y-2)2 25,过点(1,0)做两条互相垂直的弦
已知|2-x|与|x-y+4|互为相反数,求代数式3(x-y)-5(x-y)-3(x-y)+(y-x)-4(x+y)+3

|2-x|+|x-y+4|=02-x=0,x-y+4=0x=2,2-y+4=0,y=63(x-y)-5(x-y)^2-3(x-y)+(y-x)^2+4(x+y)^2+3(y-x)=-4(x-y)^2+

已知x-3y=0,求(2x+y)/(x^2-2xy+y^2)X(x-y)

(2x+y)/(x^2-2xy+y^2)X(x-y)=(2x+y)/(x-y)²*(x-y)=(2x+y)/(x-y)x-3y=0,x=3y(2x+y)/(x-y)=(6y+y)/(3y-y

已知 2x-3y/y等于2 求x-y/2x 求x-3y/2x+y

(2x-3y)/y=2=>2x-3y=2y=>y=2/5x(1)将(1)式代入待求式子:(x-y)/2x=(x-2/5x)/2x=3/10同理可得(x-3y)/(2x+y)=-1/12

已知x/y=2/3,则2x-y/x+2y的值是多少?

2x-y/x+2y是(2x-y)/(x+2y)吧(2x-y)/(x+2y)=(2x/y-1)/(x/y+2)=(2X2/3-1)/(2/3+2)=1/3÷8/3=1/8

已知:3x=8y.求(1)x+y/y (2)2x+3y/x-2y

3x=8yx/y=8/3(1)x+y/y=x/y+1=8/3+1=11/3(2)2x+3y/x-2y分子分母同时除以y得=(2x/y+3)/(x/y-2)=(16/3+3)/(8/3-2)=(25/3

已知圆x²+y²+x-6y+m=0与直线x+2y-3=0

设p(x1,y1),Q(x2,y2)因为OP⊥OQ所以x1*x2+y1*y2=0x²+y²+x-6y+m=0x+2y-3=0消去yx1,x2就是所得方程的解用韦达定理就可以了会了吗

已知x−2y−3

∵x−2y−3+|x+2y-5|=0,∴x−2y=3x+2y=5,解得:x=4,y=12,则yx=116.

已知X-Y/X+Y=3,求代数式2(x-y)/X+Y-3X+Y/X+Y

X+Y分之X-Y等于3x=-2yX+Y分之2(x-y)减X+Y分之3X+Y=(-x-3y)/(x+y)=1

已知x-y/x+y=3,求代数式5(x-y)/x+y-x+y/2(x-y)

因为(x-y)/(x+y)=3,则(x+y)/(x-y)=1/3则5(x-y)(x+y)-(x+y)/2(x-y)=5*3-1/(3*2)=15-1/6=89/6

已知xy=-2,x-y=3,求(x+y)(x-y)-y^2+(x-y)^2-(6x^2y-2xy^2)÷2y

(x+y)(x-y)-y^2+(x-y)^2-(6x^2y-2xy^2)÷2y=(x+y)(x-y)-y^2+(x-y)^2-2y(3x^2-xy)÷2y=(x+y)(x-y)-y^2+(x-y)^2

已知2x^2+3y^2

以下是柯西不等式:括号比较多,可能看得有点眼花:(x+2y)^2

已知x>0,y<0,化简:/3x-2y/-/3y-2x/

/3x-2y/-/3y-2x/=3x-2y-2x+3y=x+y

已知圆x^2+y^2+x-6y+m=0与直线x+2y-3

已知圆x^2+y^2+x-6y+m=0与直线x+2y-3=0相交于P、Q两点,O为坐标原点,若OP垂直于OQ,试求m的值.圆的方程x²+y²+x-6y+m=0可化为:(x+1/2)

已知x-2y/2x+3y=3/4,则x:y=

x-2y/2x+3y=3/44(x-2y)=3(2x+3y)4x-8y=6x+9y-2x=17yx/y=-17/2

已知x-2y/3y-x=2/3,则y/x的值是

(x-2y)/(3y-x)=2/33(x-2y)=2(3y-x)3x-6y=6y-2x5x=12y,代入原方程检验通过,故y/x=5/12

已知x-2y=3,则(2x+y-6)/(3x-y-9)=?

x-2y=3则:x=2y+3代入原式,原式=(4y+6+y-6)/(6y+9-y-9)=5y/5y=1

已知3x^2+2y^2

这个当然可以用椭圆,柯西不等式也行1*5>=(3x^2+2y^2)*(3+2)>=(3x+2y)^2-sqrt(5)

已知x+2+|y-3|=0

∵x+2≥0,|y-3|≥0且x+2+|y-3|=0,∴x+2=0,|y-3|=0,∴x+2=0,y-3=0,解得x=-2,y=3,当x=-2,y=3时,x2+y2+3=(-2)2+32+3=16=4

已知x=1/3,y=-1/2,求代数式x-(x+y)+(x+2y)-(x+3y)+(x+4y)-(x+5y)+...-(

原式=x-x+x-x+……-x+(2-1+4-3+5-4+……+2008-2007-2009)y=0+(1×1004-2009)y=-1005y=1005/2