x z＝y① 7z＝x y 2 x y z＝14③ ②

xy/(x+y)=1,取倒数(x+y)/xy=1x/xy+y/xy=11/y+1/x=1.1yz/(y+z)=2,取倒数(y+z)/yz=1/2y/yz+z/yz=1/21/z+1/y=1/2.2xz

x²+y²+z²＝xy+yz+xz两边各乘以2得到2x²+2y²+2z²=2xy+2yz+2xzx²-2xy+y²+x&

(abc)^(xyz)=a^(xyz)*b^(xyz)*c^(xyz)=[a^(yz)]^x*[b^(xz)]^y*[c^(xy)]^z=[b^(xz)]^x*[b^(xz)]^y*[b^(xz)]^

X^2+Y^2+Z^2=XY+YZ+XZ则有2X^2+2Y^2+2Z^2-2XY-2YZ-2XZ=0==>(X-Y)^2+(Y-Z)^2+(Z-X)^2=0必然X-Y=0,Y-Z=0,Z-X=0==>

结果等于：1原式=x/(xy+x+xyz)＋y/(yz+y+xyz)＋z/(xz+z+xyz)=1/(y+1+yz)+1/(z+1+xz)+1/(x+1+xy)=xyz/(y+xyz+yz)+1/(z

题目应为:xy/(x+y)=6/5yz/(y+z)=12/7xz/(x+z)=4/3求x和y和z运用倒数变形可解因为1/y+1/x=5/6,1/z+1/y=7/12,1/z+1/x=3/4三式相加得1

xy\X+Y=12\71/y+1/x=7/12(1)YZ\Y+Z=6\51/z+1/y=5/6(2)XZ\X+Z=4\31/z+1/x=3/4(3)由(1)-(2)得1/x-1/z=-1/4(4)由(

xy/(x+y)=6/5①---->(x+y)/(xy)=5/61/x+1/y=5/6(4)yz/(y+z)=12/7②1/y+1/z=7/12(5)xz/(x+z)=4/31/x+1/z=3/4(6

令x/3=y/2=z/5=k则x=3ky=2kz=5k∴(xy+yz+zx)/(x²+y²+z²)=(6+10+15)k²/(9+4+25)l²=31

VB中运算符的优先顺序是先“比较”后“逻辑”,逻辑当中又是“and”的优先级高于“or”,（xz）与（z

可以联立条件解得y=3z,x=4z,然后代入原式使用最猥琐的方法求解即可,由於是齐次式所以元一定会消掉的,结果我口算的是19分之26,不知道对不对

x-y=az-y=6所以x-z=(x-y)+(y-z)=a-6x^2+y^2+z^2-xy-yz-xz=[(x-y)^2+(y-z)^2+(z-x)^2]/2=[a^2+36+(a-6)^2]/2=a

1、本题适用的相关公式为：（a+b+c）2=a2+b2+c2+2ab+2bc+2ca∵x+y+z=6,xy+yz+xz=7∴x2+y2+z2=（x+y+z）2--(2xy+2yz+2zx)=62--2

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

（x+y+z）^2=x^2+y^2+z^2+2(xy+yz+xz)=25x^2+y^2+z^2=25-14=11

答案是：(2*X)/((X-Z)*(X+Z))再问：解题过程给我写下1再答：=（2X+Z-Y)/[(x-y)(x+z)]-(y-z)/[(x-z)(x-y)]=[(2x+z-y)(x-z)-(y-z)

令2/x=3/y=7/z=k∴x=2/ky=3/kz=7/k∴(xy+xz+yz)/(x^2+y^2+z^2)=(2/k*3/k+2/k*7/k+3/k*7/k)/(4/k²+9/k

若x与z互为相倒数,则xz=1,|y|＝7,y=±7,则xz+y＝1±7=8or-6.

x+3y+7z=02x+5y+11z=0x=2z,y=-3zz=0,x=y=0分式无意义(x^2+y^2+z^2)/(xy+2yz+3xz),z≠0=[(2z)^2+(-3z)^2+z^2]/[(-2

yz/x+xz/y≥2·√(yz/x·xz/y)=2z同理,yz/x+xy/z≥2yxz/y+xy/z≥2x相加即可得证.