3^x=7^y=63^z ,求x,y,z 的关系
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x:y:z=(3y/5):y:(7y/4)=(3/5):1:(7/4)=12:20:35再问:已知x+2y-z=02x+3y+z=0求x:y
(5x+3y+2z)+(4x+6y+7z)=2011+20129(x+y+z)=4023x+y+z=447
x/(y+z)=y/(x+z)=z/(x+y)当x+y+z=0时,x+y=-z(x+y)/z=-z/z=-1当x+y+z≠0时,由x/(y+z)=y/(x+z)=z/(x+y)根据等比性质可得(x+y
把1式减去2式得:X+3Z=43即X=43-3Z,再把X=43-3Z代入原方程式可得:Y=2Z-3把X=43-3Z,Y=2Z-3再代入原方程式可消去变量Z故此题有N个解,当Z=0时,X=43,Y=-3
答案为x=2,y=3,z=1;解答过程为:1式与3式相加可得5x+5y=25,算出x+y=5,代入2式得z=1,再把z=1代入可得x=2,y=3;要采纳哦!
给的两个等式相加7x+7y+7z=7,x+y+z=1
②-①,得x+3y=21x=21-3y③把③代入①,得3(21-3y)+7y+z=6363-9y+7y+z=63-2y+z=0z=2y④然后自己应该就知道了吧?
93x+7y+z=5所以6x+14y+2z=10又因为4x+10y+z=3所以2x+4y+z=7原题中两式相减得x+3y=-2所以x+y+z=9
x=1,y=2,z=3联立任意两个,消除z,x+2y+3z=142x+y+z=75x+y=7再联立其他的的两个.消除z2x+y+z=73x+y+2z=11x+y=3然后联立这两个x+y=35x+y=7
①代表3x+7y+z=6,②代表4x+10y+z=7,②-①得出x+3y=1,①*4,变成12x+28y+4z=24②*3,变成12x+30y+3z=21,再相减得出z-2y=3,把x+3y=1和z-
x+y+z-6=02x+3y-z-12=02x-y-z=0组成方程组再解x=2y=3z=1
x+4y+3z=3x-2y-5z=0则x+4y+3z=0①3x-2y-5z=0,则6x-4y-10z=0②①②两式相加,得7x-7z=0,所以x=z代入①,得z+4y+3z=0,所以y=-z所以x+2
3x+7y+z=5.(1)4x+10y+z=3.(2)(1)*3-(2)*2有9x+21y+3z-(8x+20y+2z)=5*3-3*2x+y+z=15-6x+y+z=9
|x-z-2|+|3x-6y-7|+(3y+3z-40)^2=0x-z-2=0,3x-6y-7=0,3y+3z-40=0x=11,y=13/3,z=9
因为|x-z-2|》=0,3x-6y-7)^2》=0,|3y+3z-4|》=0且|x-z-2|+(3x-6y-7)^2+|3y+3z-4|=0所以三个式子都等于0,得到一个方程组,解方程组得x=3,y
(4x+10y+z)-(3x+7y+z)=x+3y=105(3x+7y+z)-2(x+3y)=x+y+z=105
解法1:2x+5y+4z=0式①3x+y-7z=0式②x+y-z=?式③式①=0,式②=0,所以式①-式③=式②-式③即:2x+5y+4z-x-y+z=3x+y-7z-x-y+zx+4y+5z=2x+
两式相减(后面的减前面的)x+3y=105所以2x+6y=210第一个式子减上式x+y+z=105
①令x/2=y/4=z/7=k所以x=2k,y=4k,z=7k所以x:y:z=2:4:7②2x+y+3z=58即4k+4k+21k=5829k=58k=2x=2k=4y=4k=8z=7k=14③(x+
x-y-z=4(1)-x+3y-z=8(2)-x-y+7z=16(3)(1)(2)(3)表示1式,2式,3式,不是数字的意思(1)+(2)得:2y-2z=12(4)(1)+(3)得:-2y+6z=20