x^2(y+z)^2-2xy(x-z)(y+z)+y^2(x-z)^2 =[x(y+z)-y(x-z)]^2 =(xz+

来源：学生作业帮编辑：搜狗做题网作业帮分类：数学作业时间：2024/06/16 23:28:13

x^2(y+z)^2-2xy(x-z)(y+z)+y^2(x-z)^2 =[x(y+z)-y(x-z)]^2 =(xz+yz)^2 =z^2(x+y)^2

x²(y+z)²-2xy(x-z)(y+z)+y²(x-z)² 完全平方式分解因式
[x(y+z)]²-2×x(y+z)×y(x-z)+[y(x-z)]²
-------- ------- ------- ---------
a² -2 a b + b² =(a+b)²
=[x(y+z)-y(x-z)]² 去掉里面的括号,并合并同类项
xy+xz-xy+yz=xz+yz
=(xz+yz)² 提取公因式 [z(x+y)]²
=z²(x+y)² 公式(ab)^n=a^nb^n

已知x,y,z 大于0,x+y+z=2,求证 xz/y(y+z)+zy/x(x+y)+yx/z(z+x)大于等于2/3 (2X+Z-Y)/(X^2-XY+XZ-YZ)-(Y-Z)/(X^2-XY-XZ+YZ) 化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+（z-y)(x-y)/(xy-2z)(y+z-2x)+(x- (y-x)/(x+z-2y)(x+y-2z)+(z-y)(x-y)/(x+y-2z)(y+z-2x)+(x-z)(y-z 试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y) 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) 化简(2x-y-z/x^2-xy-xz+yz)+(2y-x-z/y^2-xy-yz+xz)+(2x-x-y/z^2-xz (x+y-z)^2-(x-y+z)^2=? 化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+（z-y)(x-y)/(x-2z+y)(y+z-2x)+(x 分解因式:f(x,y,z)=x^2(y-z)+y^2(z-x)+z^2(x-y) x,y,z为实数且(y-z)^2+(x-y)^2+(z-x)^2=(y+z-2x)^2+(x+z-2y)^2+(x+y x,y,z为实数且(y-z)平方+(x-y)平方+(z-x)平方=(y+z-2x)平方+(z+x-2y)平方+(x+y-