①﹙χ²-2y²-z²﹚-﹙﹣y²+3χ²-z²﹚+﹙5χ&
已知方程组﹛x+4y-3z=0 4x-5y+2z=0 ,xyz≠0,求﹙3x²+2xy+z²﹚÷﹙x
3(y-z)²-(2y+z)(-y+2y)
已知有理数想x,y,z满足(x-4)²+3丨x+y-z丨=0,则,(5x+3y-3z﹚的2008次方的末尾数字
①(2y-z)²[2y(z+2y)+z²]²=
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
25(x-y)²-10(y-z)+1因式分解
{x²-(y-z)²分之x²-y²}÷{(x-y)²-z²分
x²+y²+z²-2x+4y-6z+14=0,则x+y+z=?求详解
已知2分之x=3分之y=4分之z,则3x-2y+z分之2x+y-z=﹙﹚
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.
(x-3y+2z)²=
x²+2xy+y²-z² 用分组分解法因式分解