圆x平方加y平方加2x减2y等于0的半径-搜答案-大师作文网

1/2再答：再问：谢谢了，问题解决了

x^2+2x+y^2-6y+10=0=(x^2+2x+1)+(y^2-6y+9)=0=(x+1)^2+(y^2-6y+3^2)=0=(x+1)^2+(y-3)^2=0x=-1y=3xy=-3

∵(x+1)²+|y-2|=0　∴x+1=0　　y-2=0　∴x=-1　　y=2　∴x²y²+3xy-7x²y²-(5/2)xy+1+5x²

x^2-2x+y^2+6y+10=0x^2-2x+1+y^2+6y+9=0(x-1)^2+(y+3)^2=0x-1=0,y+3=0x=1,y=-3(x-y)/(x+y)=4/(-2)=-2

16x-y=2x^2+y^2=12x^3y-2x^2y^2+xy^3=xy(x^2+y^2-2xy)=xy(x-y)^2x-y=2==>(x-y)^2=4==>x^2+y^2-2xy=4又因为x^2+

前者分子：x^2-y^2=(x+y)(x-y)那前面这项就可以化简约分啦后者分子：4x^2-4xy+y^2=(2x)^2-2*2xy+y^2=(2x-y)^2建议你理解一下x^2-2xy+y^=(x-

等于负四

x平方加y平方减2xy减6x加6y加9=0x^2+y^2-2xy-6x+6y+9=0(x^2+y^2-2xy)-6(x-y)+9=0(x-y)^2-6(x-y)+9=0(x-y-3)^2=0x-y-3

两圆方程整理成标准型C1：(x-1)²+y²=1C2：x²+(y-7)²=49-a∵两圆有三条公切线∴两圆外切连心线长为两圆半径之和∵C1(1,0)C2(0,7

x^2+2xy+y^2=25===>(x+y)^2=5^2===>x+y=±5(x-y)^2-3x+3y+2=0===>(x-y)^2-3(x-y)+2=0===>[(x-y)-1]*[(x-y)-2

x²+y²+2x-8y+8=0(x²+2x+1)+(y²-8y+16)+8-17=0(x+1)²+(y-4)²=9(x+1)²+(

x的平方加2x加1减y的平方=(x+1)²-y²=(x+y+1)(x-y+1)

求根法：x^2+xy-2y^2+2x+7y-3=x^2+x(y+2)-2y^2+7y-3,△（x)=(y+2)^2-4(-2y^2+7y-3)=9y^2-24y+16=(3y-4)^2,x1=[-(y

x的平方减6xy加9y的平方减2x加6y加1=(x-3y-1)^2

x^2-4xy+4y^2-3x+6y+2=(x-2y)^2-3(x-2y)+2=(x-2y-1)(x-2y-2)

x^2+4x+y^2-2y+5=0(x^2+4x+4)+(y^2-2y+1)=0(x+2)^2+(y-1)^2=0x=-2,y=1x^2+y^2=(-2)^2+1^2=4+1=5

3∵x/y=1/2∴y=2x故（x^2+2xy+y^2)/(x^2-xy+y^2)=(x+y)^2/(x^2-xy+y^2)=(x+2x)^2/(x^2-x·2x+4x^2)=9x^2/(3x^2)=

2+x^2+(x^2)*(y^2)=2X-2xy,化简得：（x-1)^2+(xy-1)^2=0【把化简前式子中的2分成1+1】所以x-1=0,xy-1=0,得x=y=1【（x-1)^2是（x-1)的平

解题思路：此题考查了直线与圆相交的性质，涉及知识有，圆的标准方程，点到直线的距离公式.解题过程：附件

y=3x的平方减x加2

y=3x^2-x+2--->3x^2-x+(2-y)=0判别式不小于0,故(-1)^2-4×3×(2-y)>=0--->-23+12y>=0--->y>=23/12故函数y的值域为[23/12,+无穷