初二数学题 (1) (x-y-2)² (2)﹙2x+y+z﹚﹙2x-y-z﹚
有关分式的数学题1、已知X/3=Y/5=Z/4,求(X+Y+Z)/Y和(X+Y+Z)/(Y-Z)的值.2、甲乙两人分别从
x,y,z为正实数 x/(2x+y+z)+y/(x+2y+z)+z/(x+y+2z)
已知有理数x,y,z,且|x-3|+2|y+1|+(2z+1)²=0,求x+y+z的相反数的倒数.
(y-x)/(x+z-2y)(x+y-2z)+(z-y)(x-y)/(x+y-2z)(y+z-2x)+(x-z)(y-z
两道初一数学题(x+3y-2z)(x+3y+2z)-(x-2y+3z)(x+2y+3z) (5x-3y)(3y+5x)-
4x-3y-6z=0,x+2y-7z=0,xyz的积不等于0,求(x+y-z)/(x-y+2z)的值
已知x::y:z=3:4:5,(1)求x+y分之z的值;(2)若x+y+z=6,求x,y,z.
已知x/2=y/3=z/4,求下列各式 (1)(x+y+z)/x (2)(x-y+2x/(x-y-2z)
2.若实数X,Y,Z满足√X+√(Y-1)+√(Z-2)=1/2(X+Y+Z),求logz(X+Y)的值.
x,y,z正整数 x>y>z证明 x^2x +y^2y+z^2z>x^(y+z)*y^(x+z)*z^(x+y)
已知×=负二分之一y=负三分之一z=六分之一求19(x-y+z)的平方减﹣24(x-y-z﹚减112﹙x-y+z﹚的值
化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+(z-y)(x-y)/(x-2z+y)(y+z-2x)+(x