已知y=3分之1x+1求[1]当x=-1时的函数值-搜答案-大师作文网

x-3y=0x=3y(2x+y)/(x^2-2xy+y^2)÷1/(x-y)=(2x+y)/(x-y)^2÷1/(x-y)=(2x+y)/(x-y)^2*(x-y)=(2x+y)/(x-y)=(2*3

x=1/3y=1/2x-(x+y）+（x+2y)-(x+3y)+...-(x+13y)=6y-13y=-7y=-7/2=-3.5

1/x-1/y=2001则(x-xy-y)/(x-y)=2002/2001

令2分之x=3分之y=4分之z=k,则：x=2ky=3kz=4kx+y+z=1/12,得：2k+3k+4k=1/129k=1/12k=1/108所以：x=2/108=1/54y=3/108=1/36z

∵（2x+y)/2=(5x+2y)/4=1∴{2x+y=2(1){5x+2y=4(2)(1)==>y=2-2x代入（2）5x+2(2-2x)=4x+4=4,∴x=0,y=2∴(x+2y+1)/(2x-

2分之3x+y=14分之5x-y=1,3x+y=2(1)5x-y=14(2)由(1)+(2)式得8x=16x=2y=-42x-3y+7=2*2-3*(-7)+7=32

用x分之3加y分之4乘以（x+y),然后化简,再用均值不等式就出来了……(3/x+4/y)乘（x+y）等于7+3y/x+4x/y然后由于x,y大于0,所以用均值不等式即可……

1.令x/2=y/3=z/4=k,则有:x=2ky=3kz=4k所以:(2x+y-z)/(x+y+z)=(2×2k+3k-4k)/(2k+3k+4k)=1/32因为x^2+y^2+2x+4y+4=0即

3+9+3=15==

x-1的绝对值+y-3的绝对值=0,绝对值都大于等于0相加为0则各项均为0所以x-1=0y-3=0所以x=1,y=3所以xy/x+y=3/4

∵x/4=y/5　∴x:y=4:5　∵y/5=z/3　∴y:z=5:3　∴x:y:z=4:5:3

由于|x+y|≠x+y,说明x+y0则x+y=-12又4分之3+8又2分之1=-51/4+17/2=-17/4即负4又四分之一2.若y

ay/(y-b)

x+y=10x-y=8/7原式=-7³/8*(x+y)³*(x-y)*4(x+y)²(x-u)²=-343/2*(x+y)^5*(x-y)³=-343

既然｛x-2｝+｛y-三分之一｝=0,那么x=2,y=三分之一..这点懂?那么2x+3y=2×2+3×三分之一=5

(3x-4y)÷(2x+y)=1/2所以2x+y=2(3x-4y)=6x-8y4x=9y所以x/y=9/4

因为x＋y/2=3x＋4y/5=1,所以x＋y=2,3x＋4y=5所以解方程组得x=3,带入x＋5y＋6/3x－y＋1,得4/11

x=2y/3(4y/3+y-1)/2=(-y/3)/34y+3y-3=-2y/37y-3=-2y/321y-9=-2yy=9/23

原式等价于10(x+1)=5(y+3)=4(x+y)10(x+1)=5(y+3)可得2x-y=15(y+3)=4(x+y)可得4x-y=15解得x=7y=13x+2y+3=363x+2y+1=48两式