已知x.y.z满足xy x y=-2.yz y z

X+Y-Z=6①Y+Z-X=2②Z+X-y=0③①+②+③得x+y+z=8④④-①得2z=2z=1④-②得2x=6x=3④-③得2y=8y=4即x=3y=4z=1

x=1,y=2,z=310,11,12,13,21,31Aborrow从..借,他应该是向妈妈借钱lent借给有yesterday用过去式还有mothe是mother吧?

x+2y-z=6①x-y+2z=3②，①×2+②，得x+y=5，则y=5-x③，①+2×②，得x+z=4，则z=4-x④，把③④代入x2+y2+z2得，x2+（5-x）2+（4-x）2=3x2-18x

实数x,y,z,满足那么x+y=6,z^2=xy-9,∴xy=z^+9,(x-y)^=(x+y)^-4xy=-4z^>=0,∴z=0,(x+y)^z=6^0=1.

因为x/y+z+y/z+x+z/x+y=1所以x/y+z=1-y/z+x-z/x+y,两边同乘以x得x^2/y+z=x-xy/z+x-xz/x+y同理y^2/x+z=y-xy/z+y-yz/x+y,z

x+y-z=6y+z-x=2z+x-y=0三式相加得x+y+z=8-得2z=2z=1-得2x=6x=3-得2y=8y=4x=3y=4z=1

这么简单的题目,你们不要老是依靠答案,要自己算出答案来,就算错了,那也是你自己算出来的,就算你骗了老师,但你同事也骗了你自己

已知x,y,z满足y+z/x=z+x/y=x+y/z=k,求k的值解:∵y+z/x=z+x/y=x+y/z=k∴y+z/x=kz+x/y=kx+y/z=k∴kx=y+zky=x+zkz=x+y∴kx+

xyz＝x＋y＋z＜3z∴xy＜3由于x＜y,故xy＝2,x＝1,y＝2∴z＝3

1.z²-z+1/4=(z-1/2)².绝对值、根号、平方数都是非负的,而相加为0.所以都为0.即x=y,2y=z,z=1/2.所以x=y=1/4,z=1/2.2.2002x200

第一题：2x-3y=8①3y+2z=0②x-z=-2③由①+②得到：2x+2z=8④由③式得到x=z-2,带入④式得到：z=3然后解得：x=1、y=-2、z=3,那么xyz=-6第二题：由①-2②,③

x+y-z/z=y+z-x/x=z+x-y/y,应用等比定理,得（x+y-z+y+z-x+z+x-y)/(x+y+z)=(x+y-z)/z,所以(x+y+z)/(x+y+z)=（x+y-z)/z,即1

x/(y+z)+y/(z+x)+z/(x+y)=1所以x/(y+z)=1-[y/(z+x)+z/(x+y)]y/(z+x)=1-[x/(y+z)+z/(x+y)]z/(x+y)=1-[x/(y+z)+

等于0.x/(y+z)=1-[y/(z+x)+z/(x+y)]y/(z+x)=1-[x/(y+z)+z/(x+y)]z/(x+y)=1-[x/(y+z)+y/(z+x)]x2/(y+z)+y2/(z+

k(x+y+z)=2(x+y+z)算到这一步时,要讨论当x+y+z不等于0时,k=2当x+y+z=0时,y+z=-x,y+z/x=-1另一个值是-1

x+y-z=6标记为A,y+z-x=2标记为B,z+x-y=0标记为C,A+B：2Y=8Y=4A+C：2X=6X=3B+C：2Z=2Z=1然后就是结论.回答完毕.最完整的~

x+2y-z=21①x-y+2z=12②①*2+②=3x+3y=54即x+y=18得出y=18-x代入②得x+z=15得出z=15-x代入186/x²+y²+z²得出18

X+Y-Z=6.aY+Z-X=2.bZ+X-Y=0.ca,b,c三式相加X+Y+Z=8.dd式-a式2Z=2Z=1d式-b式2X=6X=3d式-C式2Y=8Y=4

括号是什么意思?只有一半

xy/(x+y)=-2(x+y)/xy=-1/21/y+1/x=-1/2yz/(y+z)=4/3(y+z)/yz=3/41/z+1/y=3/4zx/(z+x)=-4/3(z+x)/zx=-3/41/x