(2) (5y 1) (1-y)= (9y 1) (1-3y);
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设,y=k/x+a/x,然后分别把两个值带入,解方程组k=xy=-2,和交点带入第二个式子求m,再联立解方程组.
∵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=
,y1与x-1成正比例,y2与x正比例可设y1=k(x-1),y2=tx则y=y1-y2=k(x-1)-tx当X=2时,y=4;当x=-1时,y=-5则4=k-2t-5=-2k+t解得k=2,t=-1
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.
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
设y1=k1x,y2=k2(x-2),则y=k1x−k2(x−2)根据题意,得k1+k2=−1k13−k2=5解得:k1=3,k2=-4∴y=3x+4x−8当x=5时,y=35+20−8=1235.
因为:一次函数y=-1/3x+2过点(-5,y1),(3,y2)所以:y1=-1/3×(-5)+2;y2=-1/3×3+2=5/3+2=(-1)+2=11/3=1即:y1=11/3;y2=1所以:y1
y1与x成反比例:y1=n/xy2与x-2成正比例:y2=m(x-2)并且当x=3时,y=5:y=y1-y2=n/x-m(x-2)=n/3-m(3-2)=n/3-m=5当x=1时,y=-1:y=y1-
根据题意:y=K1x+K2/x,将x、Y值代入求出K1、K2再代入x=5和K1、K2值到上面方程中可解此题就是要分清正、反比例函数概念,不困难
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
不难,初中生吧?设y1=k1/xy2=k2(x-2)x=1时y=y1-y2=k1+k2=-1x=3时y=y1-y2=k1/3-k2解得,k1=3,k2=-4;故有y=3/x+4x+8
Y1与X2成正比例?还是Y1与X^2成正比例?还是还是Y1与2X成正比例?Y1=AX^2Y2=B*X^-2Y=Y1+Y2=AX^2+B*X^(-2)当X=1时,Y=5Y=AX^2+B*X^(-2)=A
yi=a*x^2,y2=b*x^(-2),a+b=5;a+b=11,显然出的题有毛病再问:如果出错了,请将其做以适当的改正,并作出正确的解答。
y=ax^2+b/(x^1/2)5=a+b18=16a+b/2解得a=1,b=4.y=x^2+4/(x^1/2)
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(
设y1=k1x2(k1≠0),y2=k2(x+2)(k2≠0).∵y=y1+y2,∴y=k1x2+k2(x+2).又∵当x=1时,y=9;当x=-1时,y=5,∴k1+3k2=9k1+k2=5,解得k
设K为y1比列系数,u为y2比列系数y1=kxy2=u1/xy=kx+u1/x把X=1时,Y=3;当X=-5时,Y=2带代入方程,就行联立求出K,U听简单的,2元一次方程,解得U=7.5K=-4.5所
设y1=kx2,y2=a(x-2),则y=kx2+a(x-2),把x=1,y=5和x=-1,y=11代入得:k−a=5k−3a=11,k=-3,a=2,∴y与x之间的函数表达式是y=-3x2+2(x-
y1=k1(x-2)...(1)y2=k2/(5x)...(2)y=2y1+y2=2k1(x-2)+k2/(5x)...(3)当x=2时,y=9/10得9/10=2k1(2-2)+k2/(5*2)即k
y1=y22x+1=5x+16-3x=15x=-5y1>y22x+1>5x+16-3x>15x>-5y1