设y>x>z,原来三位数100z+10y+x或100x+10y+z.第一种情况
设y>x>z,原来三位数100z+10y+x或100x+10y+z.第一种情况
解三元一次方程组,100x+10y+z=27(x+y+z) {x+y-z=1 (100z+10y+x)-(100x+10
(x+y-z)(x-y+z)=
若{x+3y+10z=0 则 (x+y-z)/(x-y+z)
解一道三元一次方程x+y+z=13,y-2=z,100z+10y+x+99=100x+10z+y.
设x、y、z为整数,证明:x^4*(y-z)+y^4*(z-x)+z^4*(x-y)/(y+z)^2+(z+x)^2+(
有一个三位数,百位数字为x,十位数字为y,个位数字为z,这个三位数用代数式表示为100x+10y+z 如果
100x+10y+z=27(x+y+z) x+z=y+1 100z+10y+x-100x-10y-z=99 解方程组
100x+10y+z=48(x+Y+Z) x+Z=y+3 100x+10y+z-(100z+10y+x)=198 怎么解
(y-x)/(x+z-2y)(x+y-2z)+(z-y)(x-y)/(x+y-2z)(y+z-2x)+(x-z)(y-z
1.设X ,Y,Z 成等差数列,代数式(X-Z)*(X-Z)+ 4(X-Y)(Z-Y)=
设z=ln(x^z×y^x),求dz