(x加2)的平方加(y减6)的平方加(z减1)的平方等于25的几何意义
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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
x^2+2x+y^2-6y+10=0可以化解为(x+1)^2+(y-3)^2=0,等价于x+1=0且y-3=0,所以有x=-1,y=3,故x^3*y=-3.
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^2-2x+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
3x^2+2xy+y^2-6x-6y+9=0(x+y)^2-6(x+y)+9=0(x+y-3)^2=0x+y=3
x²+y²+4x-6y+13=0(x²+4x+4)+(y²-6y+9)=0(x+2)²+(y-3)²=0x+2=0y-3=0x=-2y=3当
x^2+y^2+2x-6y+10=0(x^2+2x+1)+(y^2-6y+9)=0(x+1)^2+(y-3)^2=0平方大于等于0,相加等于0,若有一个大于0,则另一个小于0,不成立所以两个都等于0所
原等式可化为:x²-2xy+y²+6(x-y)+9+y²=0即:(x-y)²+6(x-y)+9+y²=0(x-y+3)²+y²=0
x=1,y=-3再问:过程再答:化简一下就得到方程(x-1)2+(y+3)2=0.这里的2表示平方啊,不知道怎么打平方。很明显只有x=1,y=-3才满足啊。
x的平方减6xy加9y的平方减2x加6y加1=(x-3y-1)^2
问题求什么?再问:求x加y的值再答:x^2-2x+1+y^2-6y+9=0(x-1)^2+(y-3)^2=0x=1y=32x+3y=11
x+y-2=x^2+axy+by^2-5x+y+6x+y-2是x^2+axy+by^2-5x+y+6的因式当x+y-2=0时,x^2+axy+by^2-5x+y+6=0令x=y=1,x+y-2=0x^
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
x²+y²+z²-2x-4y-6z+14=0(x-1)²+(y-2)²+(z-3)²=0因为每个平方数都≥0,它们之和为0,那么它们每个都为
由题意可得x²+y²+2x+6y+10=0(x+1)²+(y+3)²=0所以x=-1,y=-3
原式=[(2x+y)-3]²=(2x+y-3)²