6x加5y-z
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(1)2x+3y-4z=-5(2)x+y+z=6(两边同时×33x+3y+3z=18(与(1)相减得(5)(3)x-y+3z=10(与(2)相加得(4))(4)2x+4z=16(5)x+7z=23(两
根据题意知:X/3=Y/4=Z/5设X/3=Y/4=Z/5=K则X=3K,Y=4K,Z=5K所以(X+Y)/(X-2Y+3Z)=(3K+4K)/(3K-2*4K+3*5K)=7K/(10K)=7/10
(x+y)/2=(y+z)/3=(z+x)/4整理3x+3y=2y+2z4x+4y=2x+2z4y+4z=3z+3x化简的y=1/3xz=5/3x代入2x+4y+6z=120得x=9因而求的y=3z=
x+y+z=A(3x+7y+z)+B(4x+10y+z)易知3A+4B=1,7A+10B=1,A+B=1,解得A=3,B=-2,所以原式=3*5-2*6=3,以后也是这样做就好了!给分吧!
(5x+3y+2z)+(4x+6y+7z)=2011+20129(x+y+z)=4023x+y+z=447
14=1+4+9所以(x²-2x+1)+(y²+4y+4)+(z²+6z+9)=0(x-1)²+(y+2)²+(z+3)²=0平方大于等于0
原题:x,y,z满足等式(3x+4y)/2=(6x+5y)/3=(4x-y+13z)/2,那么(x+2y+z)/(2x+3y-z)的值是多少由(3x+4y)/2=(6x+5y)/3求得y=3/2x=1
x=19/14y=-3/4z=13/4再问:求过程再答:程式一:x/2+y=z/2程式二:x-y/3=5程式三:x+y-z/2=33元一次方程从程式一得到x=z-2y把x=z-2y放入程式2和程式3就
2x+5y+3z=0,3x-5y-8z=0,两式相加:5x-5z=0,所以,x:z=1(1)式乘以3:6x+15y+9z=0(2)式乘以2:6x-10y-16z=0两式相减:25y+25z=0,y:z
(1)|x-z-2|+(3x-6y-7)^2+|3y+3z-4|=0所以:x-z-2=03x-6y-7=03y+3z-4=0所以解方程组:x=3;y=1/3;z=1(2)设甲,乙,丙配成一套的比例为:
X等于51/14Y等于1/7Z等于73/14
x²+y²+z²-2x-4y-6z+14=0(x-1)²+(y-2)²+(z-3)²=0因为每个平方数都≥0,它们之和为0,那么它们每个都为
x/3=y/4=z/5x=3/4y,z=5/4y(x+y-z)/(x-y+z)=(3/4y+y-5/4y)/(3/4y-y+5/4y)=1/2y/y=1/2
x/z+z/x(4y)^2=3x*5z2/y=1/x+1/zy=2xz/(x+z)(4y)^2=[8xz/(x+z)]^2=15xz64xz=15(x+z)^2,除以z^264x/z=15(x/z+1
设2/x=3/y=5/z=1/t∴x=2t,y=3t,z=5t∵2x+y-z=4∴4t+3t-5t=4∴t=2∴x-3y+2z=4-18+20=6再问:不用举例法呢?再答:举例法?再问:不用举例法再答
(x+y+z)^5-(x+y-z)^5-(x+z-y)^5-(z+y-x)^5=80xyz(x^2+y^2+z^2)注:x^5,y^5,z^5之类的是被消掉了.我的结果100%是正确的,你再算算吧.朝
x+y=z
x-y+z=02x+y-2z=6x=z-y2x=6-y+2z代入2(z-y)=6-y+2z2z-2y=6-y+2z-2y=6-yy=-6再答:后面自己应该能算了吧,手机打字真辛苦再问:嗯谢谢你了再问:
x-y=6(1)2x+3y=z(2)3x+2y-3z=2(3)(2)代入(3)消去Z得:3X+2Y-6X-9Y=2-3X-7Y=23X+7Y=-2(4)(4)和(1)组成二元一次方程组(4)-(1)乘
应该是3X=4Y,5Y=6Z吧?X+Y:Y+Z=[(4Y/3)+Y]:(Y+5Y/6)=14;11