将(x+y-z)²(x+y+z)-(y+z-x)(z-x-y)²用提公因式法分解因式.
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/08/13 09:40:06
将(x+y-z)²(x+y+z)-(y+z-x)(z-x-y)²用提公因式法分解因式.
![将(x+y-z)²(x+y+z)-(y+z-x)(z-x-y)²用提公因式法分解因式.](/uploads/image/z/17801791-7-1.jpg?t=%E5%B0%86%28x%2By-z%29%26sup2%3B%EF%BC%88x%2By%2Bz%29-%28y%2Bz-x%29%28z-x-y%29%26sup2%3B%E7%94%A8%E6%8F%90%E5%85%AC%E5%9B%A0%E5%BC%8F%E6%B3%95%E5%88%86%E8%A7%A3%E5%9B%A0%E5%BC%8F.)
提公因式=(z-x-y)² (x+y-z)²=(z-x-y)²
(x+y-z)²(x+y+z-y-z+x)=(x+y-z)²*2x
(x+y-z)²(x+y+z-y-z+x)=(x+y-z)²*2x
分解因式 x²(x²-y²)+z²(y²-x²) (a+b)
分解因式:(1):x(x-y)+y(y-X) 多项式(x+y-z)(x-y+z)-(y+z-X)(Z-x-y)的公因式是
用提公因式法分解因式:(x^2-xy)+z(x-y)
球一到数学题:分解因式:x^4+y^4+z^4-2x²y²-2x²z²+2y&s
数学 多项式(x+y-z)(x-y+z)-(y+z-x)(z-x-y)公因式
分解因式x^2(y-z)+y^2(z-x)+z^2(x-y)
初一数学分解因式题x(x-y-z)-y(y-x+z)-z(z-x+y),看清符号呀~
分解因式:f(x,y,z)=x^2(y-z)+y^2(z-x)+z^2(x-y)
已知:x+y+z=1,x²+y²+z²=2,x³+y³+z³
已知x+y+z=3,x²+y²+z²=19,x³+y³+z³
1998(z-y)+1999(y-z)+2000(z-x)=0 1998²(z-y)+1999²(y
设x,y,z∈,R求证:x²+xz+z²+3y(X+y+z)≥0