Z=5X-Y 15,已知X与Y不相关
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(x+y-z)/z=(y+z-x)/x=(z+x-y)/y[x+y]/z-1=[y+z]/x-1=[z+x]/y-1[x+y]/z=[y+z]/x=[z+x]/y设[x+y]/z=[y+z]/x=[z
已知x.y.z都是不为0的有理数,且满足2x-5y+2z=0求x:y:z的值x+4y-12z=02x-5y+2z=0(1)x+4y-12z=0(2)(2)→2x+8y-24z=0(3)(1)-(3)-
x:z=5:4x:Y:Z=x:(3x/5):(4x/5)=5:3:4再问:过程,谢谢再答:4x=5z,两边同时除以z,得x:z=5:43x=5y,4x=5z,所以y=3x/5,z=4x/5,所以x:Y
(1)Y与Z的比值为1.25(2)设Y=15t,则X=25t,z=12t则x:y:z=25t:15t:12t=25:15:12再问:第一个,怎么出来的啊?再答:4Y=5Z则Y:Z=5:4=1.25
x+y-z=0①2x-y-5z=0②①+②得3x-6z=0x=2z代入①得y=z-x=-z将x=2zy=-z代入所求得(2x+y+3z)/(3x-y-z)=(4z-z+3z)/(6z+z-z)=6z/
已知x.y.z为三个非负有理数,且满足3x+2y+z=5,x+y-z=2,若s=2x+y-z,求s的最大值与最小值.3x+2y+z=5①2x+y-3z=1②①-②×2得7z-x=3∴z=(x+3)/7
2x-3y-4z=01式x+y+z=02式1式+2式×4得到:2x-3y-4z+4x+4y+4z=06x+y=06x=-yx:y=(-1):61式-2式×2得到:2x-3y-4z-2x-2y-2z=0
我来解答前面的解法吧把两式编号为1和2,(1)3x-2y-5z=0,(2)2x-5y+4z=0.将(1)式乘以2,得6x-4y-10z=0(3);将(2)式乘以3,得6x-15y+12z=0(4);(
答案为1.方程组中两个方程式相加,消去y,得3x-6z=0,故x=2z.代入所求式得(y+7z)/(5z-y)=A.再用方程组中第一个方程式乘以2,减去第二个方程式消去x得3y+3z=0,故y=-z代
由题意得:x+y=3①y+z=4②x+z=5③①+②+③得:2x+2y+2z=12,即x+y+z=5.故选A.
由x+y-z=02x-y-5z=0得到x=2z,y=-z所以(2x+y+3z)/(3x-y-z)=(6z)/(6z)=1
【解】视z为常数,由已知两方程,可解得x=3zy=2z将其代入待求值式中,得3x*x+2y*y+5z*z/5x*x+y*y-9z*z=[3(3z)^2+2(2z)^2+5z^2]/[5(3z)^2+(
【解】视z为常数,由已知两方程,可解得x=3zy=2z将其代入待求值式中,得3x*x+2y*y+4z*z/5x*x+y*y-9z*z=[3(3z)^2+2(2z)^2+4z^2]/[5(3z)^2+(
3x-2y-5z=012x-8y-20z=02x-5y+4z=010x-25y+20z=0两式相加22x-33y=022x=33yy=2/3x3x-2y-5z=015x-10y-25z=02x-5y+
设x+y-z/z=x-y+z/y=y+z-x/x=k有x+y-z=kzx-y+z=kyy+z-x=kx三式相加得x+y+z=k(x+y+z)k=1得x+y=(k+1)zx+z=(k+1)yy+z=(k
2x-3y-z=0(1)x+3y-14z=0(2)(1)+(2)3x-15z=0x=5z(2)*2-(1)6y-28z+3y+z=09y=27zy=3z代入(4x^2-5xy+z^2)/(xy+yz+
256X的四次方y的六次方-81y的二次方z的四次方=y(256x^4y^4-81z^4)=y(16xy+9z)(16xy-9z)=y(16xy+9z)(4xy+3z)(4xy
4x-3y-3z=0①x-3y+z=0②①-②,得3x-4z=03x=4z由于z不等于0,故有x:z=4:3同理可得:①-4②,得9y-7z=09y=7zy:z=7:9
应该是3X=4Y,5Y=6Z吧?X+Y:Y+Z=[(4Y/3)+Y]:(Y+5Y/6)=14;11