2y+z=3y ① ,x+3y-z=8,② x+y+2z=13③ 求x,y,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) 2y+z=3y ① ,x+3y-z=8,② x+y+2z=13③ 求x,y,z 已知x-y/x+y=y+z/2(y-z)=z+x/3(z-x),求证8x+9y+5z=0THX.. (4x-2y-z)-{5x-[8y-2z-(x-2y)]-x-(3y-10z)}=? {x+y+z=1;x+3y+7z=-1;z+5y+8z=-2 (4x-2y-z)-{5x-[8y-2z-(x+y)]-2(3y-10z)}=?好难啊 证明 ：x/(y+z)+y/(z+x)+z/(x+y)>=3/2其中 x,y,z>0 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=? x+2y=3x+2z=4y+z 求x:y:z x+y+z=6 x+2y+3z=14 y+1=z 急 x,y,z均为正整数,x>=y,y>=z,z>=8 x+3y-z=急x,y,z均为正整数,x>=y,y>=z,z>=8x+3y-z=132,请问1.x+y+z有最小值782.x+y+z有最大值1343.洽有一组解(x,y,z)使得x+y+z有最大值4.x+3y有最小值1405.x+2y有最大值130 (4x-2y-z)-{5x[8y-2y-(x+y)]-x+(3y-10z)]=? kuaihuajian ①x+y+z=6 3x-y+2z=12 x-y-3z=-4 ②x+y-z=2 4x-2y+3y+8=0 x+3y-2z-6=0③x/3=y/2 y/4=z/5 x+y+z=60三元一次方程 已知实数x y z满足x/(x+1)=y/(y+2)=z/(z+3)=(x+y+z)/3,求x+y+z的值