求证:(x^2-xy+y^2)^3+(x^2+xy+y^2)^3能被2x^2+2y^2整除
来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/11/14 07:47:30
求证:(x^2-xy+y^2)^3+(x^2+xy+y^2)^3能被2x^2+2y^2整除
有公式
a³+b³=(a+b)(a²-ab+b²)
故
(x^2-xy+y^2)^3+(x^2+xy+y^2)^3
=(x^2-xy+y^2+x^2+xy+y^2)((x^2+xy+y^2)^2-(x^2-xy+y^2)*(x^2+xy+y^2)+(x^2+xy-y^2)^2)
=(2x^2+2y^2)((x^2+xy+y^2)^2-(x^2-xy+y^2)*(x^2+xy+y^2)+(x^2+xy-y^2)^2)
有因式2x^2+2y^2
因此
(x^2-xy+y^2)^3+(x^2+xy+y^2)^3能被2x^2+2y^2整除
a³+b³=(a+b)(a²-ab+b²)
故
(x^2-xy+y^2)^3+(x^2+xy+y^2)^3
=(x^2-xy+y^2+x^2+xy+y^2)((x^2+xy+y^2)^2-(x^2-xy+y^2)*(x^2+xy+y^2)+(x^2+xy-y^2)^2)
=(2x^2+2y^2)((x^2+xy+y^2)^2-(x^2-xy+y^2)*(x^2+xy+y^2)+(x^2+xy-y^2)^2)
有因式2x^2+2y^2
因此
(x^2-xy+y^2)^3+(x^2+xy+y^2)^3能被2x^2+2y^2整除
(x-2xy)*(-xy+2y*y)-(3x*x-2xy)(x-9xy+6y*y)
(xy-x^2)乘以(xy)/(x-y)
(x+y)(x-y)-(2xy+3y)平方
x^2-xy-6y^2+3xy+y+2
(x^2+xy/x-y)/(xy/x-y)计算
化简:x-y/x+3y÷x^-y^2/x^2+6xy+9y^2-(xy/x+y)
1、(-7x^y)(2x^y-3xy^3+xy) 2、((x-y)^6)/((y-x)^3)/(x-y)
[x(x^2y^2-xy)-y(x^2-x^3y)]/3x^2y
计算3xy[2xy-x(y-2)+x-1]
计算:3xy[2xy-x(y-2)+x]
(-3x^y+2xy)-( )=4x^+xy
(3xy)平方/(-xy)+(x-2y)平方-(x+2y)(x-2y)=