已知y等于p加z
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像这种题就可以把未知数化成一个来解,再代入接其他的未知数X:Y:Z=3:4:5因此Y=4/3XZ=5/3X已知2X-3Y+2Z=8即2X-3*4/3X+2*5/3X=8求出:X=6故Y=8Z=10X+
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=
因为根式和绝对值都>=03x-5y=0,x-3y+3=0y=9/4,x=15/42x-2/3y=30/4-3/2=6平方根是根号6,和-根号6\x0d很不错哦,你可以试下g┷hz∠r摔be狮カwбl畅
3分之14再答:将两个方程相加,则去掉3y,得6x+6z=28两边同时除以6得x+z=3分之14
3×5=15所以2^x×2^y=2^z2^(x+y)=2^z所以x+y=z
x/2=y/3=z/4=tx=2ty=3tz=4tx+y-z=1/122t+3t-4t=t=1/12x=1/6y=1/4z=1/3xyz=1/6*1/4/*1/3=1/72
X=六分之一Y=四分之一Z=三分之一
(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:
XY/X+Y=-2,-->(x+y)/(xy)=-1/2,-->1/x+1/y=-1/2YZ/Y+Z=4/3,-->(y+z)/(yz)=3/4,-->1/y+1/z=3/4ZX/Z+X=-4/3,-
x+1/y=11/y=1-xy=1/(1-x)z+1/x=1z=1-1/x=(x-1)/x1/z=x/(x-1)y+1/z=1/(1-x)+x/(x-1)=(1-x)/(1-x)=1
x^2-z^2=(x-z)*(x+z)x+z=14x-y=2→y=x-2y-z=2→x-2-z=2→x-z=4所以x^2-z^2=(x-z)*(x+z)=14*4=56
x^2+y^2+z^2-2x+4y-6z+14=0(x^2-2x+1)+(y^2+4y+4)+(z^2-6z+9)=0(x-1)^2+(y+2)^2+(z-3)^2=0x=1,y=-2,z=3(x-y
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-y=5,y-z=3,∴x-z=8由x-y-5,得:x2-2xy+y2=25同理,可得:y2-2yz+z2=9x2-2xz+z2=64三式相加,得:2x2+2y2+2z2-2xy-2yx-2xz=
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
设:a减b分之x等于b减c分之y等于c减a分之z=K则X+Y+Z=ak-bk+bk-ck+ck-ak=0
题:8x+y-z=82x-y+z=18求8x+y+z原方程组变形为:8x+y=z+8……①2x-y=-z+18……②①+②得:10x=26x=13/5……③①-②×4得:5y=5z-64y=z-64/
x+y=z