已知x^2 y^2-2x-6y 10=0
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(1)y2=x^2+6x+12-ax^2+2akx-ak^2-2……①带入x=k,和y2=17,得k^2+6k-7=0,由k〉0得k=1(2)y2=x^2+6x+12-ax^2+2akx-ak^2-2
∵y1与3x成正比例y2与(x+5)成正比例∵y1=k1×3xy2=k2(x+5)(k1k2≠0)∵y=2y1-y2=6k1-k2(x+5)∴12=6k1-6k2-2=-6k1-4k2∴k1=1k2=
由{y=1/2x-2{y=-x+1可得1/2x-2=-x+1即x=2则y=-1所以{y=1/2x-2{y=-x+1的解是{x=2{y=-1画图后可看出:当x>2时,y1>y2当x<2时,y1
y1=k1x,y2=k2/x^2y=k1x+k2/x^2代入数值-4=k1+k25=2k1+k2/4k1=24/7,k2=-52/7y=24/7x-52/7x^2
两式相减得:2x(x2-x1)+2y(y2-y1)+x1^2-x2^2+y1^2-y2^2=A-B得:y=kx+t,这里k=(x1-x2)/(y2-y1),t=(A-B-x1^2+x2^2-y1^2+
设Y1=K1X,Y2=K2X^2∴Y=3K1X-K2X^2,得方程组:-2=6K1-4K2-9=9K1-9K2化简得:3K1-2K2=-1K1-K2=-1解得:K1=1,K2=2∴Y=3X-2X^2再
设y2=ax把(2,2)代入得a=1y2=x设y1=b/x把(2,4)代入得b=8y1=8/x当x=√3时,y1=√3y2=8√3/3
y1与x^2成反比例,y1=m/x^2y2与x+2成正比例,y2=n(x+2)y=y1+y2=m/x^2+n(x+2)x=1时,y=9;x=-1时,y=59=m/1+n(1+2)=m+3n.(1)5=
1.设Y1=AXY2=B/(X-1)则Y=AX^2+B/(X-1)把当x=-1时,y=3;当x=2时,y=-3代入,得;y=-9/2x^2+15/(X-1)然后,把X的值代入就可以得到函数y的值了.2
1)设y1=k1/(3x),y2=k2(x)^2,则:y=y1-y2=k1/(3x)-k2(x)^2当x=1时,y=6,6=(k1/3)-(k2)当x=-1时y=-2,-2=(-k1/3)-(k2)解
y=2y1-3y2(1)依题意:设y1=k1x,y2=k2/x∴y=2y1-3y2=2k1x-3*k2/x……①将x=1代入①得:y=2k1-3k2=1将x=2代入②得:y=4k1-3k2/2=5化简
不妨设y1=k1/x,y2=k2*(x-1);则y=(2*k1)/x-k2*(x-1);将(2,3)、(-1、-6)代入,得:k1-k2=3-2*k1+7*k2=-6联立上式得:k1=3k2=0综上,
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因为y1与(x-1)成正比例,y2与x成反比例,所以y1=k1(x-1),y2=k2/x所以y=y1+y2=k1(x-1)+k2/x因为当x=2时,y=1;当x=-2时,y=-2所以k1+k2/2=1
x1=2cosay1=4sina设那点是Q则A(2cosa+4sina,2cosa-4sina)x=2cosa+4sinay=2cosa-4sina所以x+y=4cosax-y=8sinasin&su
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设y1=k1*√x,y2=k2*(1/x^2);x=1时,y=y1+y2=k1+k2=11x=4时,y=y1+y2=2*k1+k2/16=6.5解得k1=3,k2=8y=3*√x+8/x^2x=9时,
已知y=2y1-y2,y1与x成反比例,y2与(x-1)成正比例,当x=2时y=3;x=-2时y=-6,求y与x之间的函数解析式y=2a/x-bx+b当x=2时y=3;x=-2时y=-6,a-b=3-
设y1=kx,y2=m/x,则k+m=6,2k+m/2=6,所以k=2,m=4,所以x=-4时,y的值是-8-1=-9
Y2>Y3>Y1a=0ora^2+2a+1-4a=0a=1