求抛物线y等于x的平方减6x加5-搜答案-大师作文网

应该是x^2-2x+y^2+6y+10=0配方：x^2-2x+1+y^2+6y+9=0(x-1)^2+(y+3)^2=0得x=1y=-3所以x+y=-2

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

再答：谢谢

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^2+y^2-6x+4y+13=0(x-3)^2+(y+2)^2=0所以x=3y=-2x^y=3^(-2)=1/9

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

x^2+y^2=6y-4x-13x^2+4x+4+y^2-6y+9=0(x+2)^2+(y-3)^2=0x=-2y=3x^2-y^2=(-2)^2-3^2=4-9=-5

X²+Y²-6X+4Y+13=0(X²-6X+9)+（Y²+4Y+4)=0(X-3)²+(Y+2)²=0X-3=0;Y+2=0X=3,Y=-

x+y=8x-y=6所以：2x=14,即x=7y=8-7=1x的平方减y的平方减2x减2y=7²-1²-2×7-2×1=32

X²-Y²=12,即(X+Y)(X-Y)=12.又X+Y=6,则X-Y=2.故(X+Y)+(X-Y)=6+2,即2X=8,X=4;所以,Y=6-X=2.

x²-y²=(x+y)(x-y)=6×5=30

再问：第2题再答：我没写吗

再答：请采纳再问：谢谢凉

令y=0则可得x²+2x-3=0,(x+3)(x-1)=0,x1=-3,x2=1,∴交点坐标为(-3,0),(1,0).

xy=0.5×【（x-y）-x-y】=1-9=-8

直线x=负1

x+y=a-----得到(x+y)²=x²+2xy+y²=a²①x-y=b-----得到(x-y)²=x²-2xy+y²=b

这应该是两个题1、已知抛物线y=x2+2m-m2即：y等于x的平方加2m减m的平方,抛物线过原点,求m的值抛物线过原点,有x=y=0所以0=0+2m-m²m(m-2)=0m=0或m=22、已