y的平方减y等于2,求y的值
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1/2再答:再问:谢谢了,问题解决了
x-y=2,两边都是平方(x-y)^2=4x^2+y^2-2xy=4-2xy=4-3=1xy=-1/2
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
x+y-2=0x+y=2x²-y²=(x+y)(x-y)=6x-y=6/(x+y)=6/2=3x-y-5=3-5=-2
是x2-xy-3y2=0么,如果是的话可以用十字相乘,得到(x-2y)x(x+y)=0所以x=2y或x=-y所以x比y等于2或-1再问:不对啊再答:我算出来是这个,我想知道这是初几还是高几的数学。。。
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.
y=6-xy²=2+7x所以36-12x+x²=2+7xx²-19x+34=0(x-2)(x-17)=0x=2,x=17y=6-x所以x=2,y=4x=17,y=-11
x²+y²-4x-2y+5=0x²-4x+4+y²-2y+1=0(x-2)²+(y-1)²=0x-2=0x=2y-1=0y=1√(x+y)/
用到求根公式再问:怎么解用一元二次方程公式法再答:再问:太给力了,你的回答完美解决了我的问题!
令a=y/xy=ax代入(1-a²)x²-4x+1=0x是实数,方程有解所以判别式大于等于016-4+4a²>=0a取任意实数所以没有最大值
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)=
X²-y²=8展开得(x+y)(x-y)=8因为x-y=2所以x+y=4联立两式解方程得x=3,y=1所以x²+y²=3²+1²=9+1=1
x-y=(x²-y²)/(x+y)=15/5=3;
x^2-y^2=(x+y)(x-y)=3(x-y)=9x-y=3
x-y=(x+y)(x-y)=13因为x+y=5所以x-y=13/5