(y 1)(2y-3)--2(y-1),其中-5y=20
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设y1=k1xy2=k2/x则y=k1x+k2/x带入(1,3)和(2,0)两点组成方程组k1+k2=3和2k1+k2/2=0解得k1=-1k2=4则y=-x+4/x你题目中的y=2应该是x=2吧如果
设y1=k1x^2,y2=k2/(x+3)y=y1-y2=k1x^2-k2/(x+3)因为-k2/3=2k2=-69k1+6/6=0k1=-1/9因此y=-x^2/9+6/(x+3),x不等于-3
-k^2-20;x>0时,yy3>y2
∵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=
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=
设y1=k1x,y2=k2(x+1),则y=k1x-2k2(x+1),根据题意得3=k1−4k25=2k1−6k2,解得:k1=1k2=−12.∴y=x-2×(-12)(x+1)=2x+1.
分母不为0所以y不等于1,当y>1时-1
x=(y-3)/(y-1)xy-x=y-3(x-1)y=x-3y=(x-3)/(x-1)设y1=kx(k<0),y2=m(x-2)(m>0)y=kx-m(x-2)代x=1,y=-1;x=3,y=5得-
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
1、设y1=kx²,y2=m(x-2)y=kx²+m(x-2)把x=-2,y=12代入得:12=4k-4m;即:k-m=3,k=m+3;把x=-1/2,y=3代入得:3=k/4-5
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,
设Y1=k1(x-1),Y2=k2(2x+3)-1=0*k1-5*k22=2*k1-9*k2得k1=1.9k2=0.2Y=1.5x-2.5
设y1=2mx,y2=n(x+1),(m,n均为常数,m≠0,n≠0),则y=2mx+n(x+1)分别把x=1,y=2;x=-2,y=3代入y=2mx+n(x+1)得关于m,n的方程组:2=2m+2n
y1与x成正比例y1=kxy2与x-1成正比例y2=m(x-1)所以y=kx+m(x-1)当x=-1时,y=2;当x=2时,y=5,所以2=-k-2m5=2k+m所以k=4,m=-3所以y=4x-3(
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-
由题意得y2=x-2将x=-3代入得y2=-54=y1+y1y2得y1=-1将x=3代入得y2=1y=-1+-1=-2
设y1=k1/x,y2=k2x那么y=k1/x+k2x把x=2时y=14,x=3时,y=281/3代进去得方程组k1/2+4k2=14k1/3+9k2=28又1/3解方程组得k1=4k2=3
Y2>Y3>Y1a=0ora^2+2a+1-4a=0a=1
已知一元二次方程y^2-3y+1=0的两个实数根y1,y2,则y1+y2=3,y1y2=1(y1-1)(y2-1)=y1y2-y1-y2+1=y1y2-(y1+y2)+1=1-3+1=-1