已知y 2等于
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/28 12:00:50
![已知y 2等于](/uploads/image/f/4230728-8-8.jpg?t=%E5%B7%B2%E7%9F%A5y+2%E7%AD%89%E4%BA%8E)
∵x2-5xy+6y2=0,∴(x-2y)(x-3y)=0∴x-2y=0,x-3y=0,即x=2y,x=3y,∴y:x等于12或13.故选C.
题目不清楚怎么写再问:已知抛物线Y^2=AX的焦点为F(1,0),A(x1,y1),B(1,y2),C(x3,y3),(0小于等于y1小于Y2小于Y3)为抛物线上的三个点,且AF的绝对值+CF的绝对值
1)设y1=k1(x+1),y2=k2(x-2)y=y1+y2=k1(x+1)+k2(x-2)=(k1+k2)x+k1-2k2x=2,y=9,则9=2(k1+k2)+k1-2k2得k1=3x=3,y=
Y1与X的平方成正比,可以设Y1=k1x^2.Y2与X成反比例,可以设Y2=k2/xY=Y1+Y2=k1x^2+k2/x也就是Y=k1x^2+k2/x把x=1代入得到k1+k2=3,把x=-1代入得到
设y1=k1x²,y2=k2(x-2)∵y=y1+y2∴y=k1x²+k2(x-2)根据题意得5=k1-k211=k1-3k2∴k1=2k2=-3∴y=2x²-3(x-2
∵x-y=9,xy=8,∴x2+y2=(x-y)2+2xy,=92+2×8,=81+16,=97.故选B.
y=y1+y2;y1=k1x;y2=k2/x;所以y=k1x+k2/x由已知条件可知:k1+k2=4and2k1+k2/2=5得知k1=k2=2所以问题1:y=2x+2/x问题2:当x=4时,y=17
首先需说明一楼|y2-y1|=k|x2-x1|为错误,因为这样k>=0,因为题目k属于R,故与题目不符!应去掉绝对值符号(y2-y1)/k=x2-x1|AB|=√(x2-x1)^2+(y2-y1)^2
y1y2相等时,两式联立,x+3=3x-4解得x=3.5可取x小于3.5的任何数,方便起见,取零,那么y1=3,y2=-4,所以y1大于y2(这一部在演草纸上进行)因为为线性函数,所以当x小于3.5时
∵x-y=5,∴x2+y2-2xy=25①,∵(x+y)2=49,∴x2+y2+2xy=49②,∴①+②得:2(x2+y2)=74,∴x2+y2=37.故答案为:37.
a·b=(x1,y1-1)·(x2,y2-2)=x1x2+(y1-1)(y2-2)=x1x2+y1y2+2-2y1-y2再问:题目打错了,是y2-1再答:a·b=(x1,y1-1)·(x2,y2-1)
∵x+y=2,∴12x2+xy+12y2=12(x2+2xy+y2)=12(x+y)2=12×22=2.故选A.
(x+y)^2=x∧2+y∧2+2xy=9x∧2+y∧2=9-2×(-9)=27
双曲线x²/9-y²/a=1的渐近线是x/3±y/√a=0即,渐近线为√ax±3y=0∵c²=9+a∴c=√(9+a)利用对称性,不妨设右焦点为F(c,0)则焦点到渐近线
∵x2-y2=(x+y)(x-y)=6,x-y=1,∴x+y=6.故选D.
6x2-xy-15y2=(2x+3y)(3x-5y)=0,所以x=-3/2y或x=5/3y
即y1=3y2-2所以3x+1=3(5x+3)-23x+1=15x+9-212x=-6x=-1/2
f(x)=y1=5-k(x-a)^2(k>0)g(x)=y2=q(x-m)^2-2(q>0)g(a)=25y1+y2=X^2+16X+13-k(x-a)^2+q(x-m)^2=x^2+16X+10x=
1.1/2(x+1)=-x+33x=5x=5/32.1/2(x+1)+(-x+3)=0x=73.1/2(x+1)-(-x+3)=-2x=11