3x+2y+z=14 ,x+y+z=10,z+2x+3y=1 5
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3x+2y+z=14 ,x+y+z=10,z+2x+3y=1 5
3x+2y+z=14 ,x+y+z=10,z+2x+3y=1 5
3x+2y+z=14 ,x+y+z=10,z+2x+3y=1 5
x=1 y=2 z=7
x+y+z=6 x+2y+3z=14 y+1=z
X/10=Y/8=Z/9,则X+Y+Z/2Y-3Z
已知(x+y)(x+z)=x,(y+z)(y+x)=2y,(z+x)(z+y)=3z,求x,y,z
试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)
[3x+2y+z=14,x+y+z=10,2x+3y-z=1]
3x+2y+z=14 { x+y+z=10 2x+3y-z=1
3x+2y+z=14 ,x+y+z=10,z+2x+3y=1 5
X=3z X+y+z=12 X+2y-2z=10
2x+z=10 x+y-z=4 3x-y-z=0
三元一次方程组,1.x-2y+z=9 2x+y+3z=10 3x+2y1.x-2y+z=92x+y+3z=103x+2y-z2.3x-5y+2z=14x+5y-z=5X-5y+3z=-23.7x+2y-z=76x-y+z=74x+3y+z=94.4x+3y-2z=-75x-6y+z=12x+z=-25.2x+3y-4z=-23x-2y+5z=262x+5z=31
(4x-2y-z)-{5x-[8y-2z-(x-2y)]-x-(3y-10z)}=?
x+2y+3z=23 y-z=5 x+2z=10
证明 :x/(y+z)+y/(z+x)+z/(x+y)>=3/2其中 x,y,z>0
x+y+z=14 7z=x+y+2 x+z=y
(4x-2y-z)-{5x-[8y-2z-(x+y)]-2(3y-10z)}=?好难啊
x/2=y/3=z/5 x+3y-z/x-3y+z
x+y/2=z+x/3=y+z/4 x+y+z=27
若x+y+z=3y=2z,则x/x+y+z=?