百度云盘在线解压:分解因式题

来源:百度文库 编辑:神马品牌网 时间:2024/05/05 19:36:25
1. (xy+1)(x+1)(y+1)+xy
2. ( x^3+y^3)^2-4xy[x^4+(xy) ^2+y^4-2xy(x^2-xy+y^2)]
3. x^3 (a+1)-xy(x-y)(a-b)+y^3(b+1)
4. (ab+cd)(a^2-b^2+c^2-d^2)+(ac+bd)(a^2+b^2-c^2-d^2)
5. xyz(x^3+y^3+z^3)-(xy) ^3-(yz) ^3-(zx) ^3
6. 已知x+y-z是复项式x^2+axy+by^2-5x+y+6的一个因式,求a、b的值并分解因式

做完了!

1.
(xy+1)(x+1)(y+1)+xy
=(xy+1)(xy+1+x+y)+xy
=(xy+1)²+(x+y)(xy+1)+xy
=(xy+1+x)(xy+1+y)

2.
(x³+y³)²-4xy[x^4+(xy)²+y^4-2xy(x²-xy+y²)]
=(x³+y³)²-4xy[(x^4+2x²y²+y^4)-x²y²-2xy(x²-xy+y²)]
=(x³+y³)²-4xy[(x²+y²)²-(xy)²-2xy(x²-xy+y²)]
=(x³+y³)²-4xy[(x²+y²+xy)(x²+y²-xy)-2xy(x²-xy+y²)]
=(x³+y³)²-4xy(x²+y²-xy)(x²+y²+xy-2xy)
=(x+y)²(x²+y²-xy)²-4xy(x²+y²-xy)²
=(x²+y²+2xy-4xy)(x²+y²-xy)²
=(x-y)²(x²+y²-xy)²

3.
x³(a+1)-xy(x-y)(a-b)+y³(b+1)
=x³(a+1)-xy(x-y)(a+1)+xy(x-y)(b+1)+y³(b+1)
=x(a+1)(x²-xy+y²)+y(b+1)(x²+y²-xy)
=(x²+y²-xy)[(a+1)x+(b+1)y]

4.
(ab+cd)(a²-b²+c²-d²)+(ac+bd)(a²+b²-c²-d²)
=(ab+cd)(a²-d²)-(ab+cd)(b²-c²)+(ac+bd)(a²-d²)+(ac+bd)(b²-c²)
=(ab+cd+ac+bd)(a²-d²)+(ac+bd-ab-cd)(b²-c²)
=(a+d)(b+c)(a+d)(a-d)+(a-d)(c-b)(b+c)(b-c)
=(a-d)(b+c)[(a+d)²-(b+c)²]
=(a-d)(b+c)(a+d+b+c)(a+d-b-c)

5.
xyz(x³+y³+z³)-x³y³-y³z³-z³x³
=x^4yz+xyz(y³+z³)-y³z³-x³(y³+z³)
=yz(x^4-y²z²)+x(yz-x²)(y³+z³)
=yz(x²+yz)(x²-yz)-x(x²-yz)(y³+z³)
=(x²-yz)(x²yz+y²z²-xy³-xz³)
=(x²-yz)[y²(z²-xy)-xz(z²-xy)]
=(x²-yz)(y²-xz)(z²-xy)

6. 已知x+y-z是复项式x^2+axy+by^2-5x+y+6的一个因式,求a、b的值并分解因式
x²+axy+by²-5x+y+6=(x-3)(x-2)+axy+y+by²。
根据十字相乘法,该式经因式分解后的形式应为(x+my-3)(x+ny-2)。
①m=1,z=3,此时对应展开式的系数得n=b,1+n=a,-2-3n=1,得n=-1,a=0,b=-1,
x²-y²-5x+y+6=(x+y-3)(x-y-2)
②n=1,z=2,此时对应展开式的系数得m=b,1+m=a,-3-2m=1,得m=-2,a=-1,b=-2,
x²-xy-2y²-5x+y+6=(x+y-2)(x-2y-3)

此类问题是你的作业吧?没有难度啊,为什么全拿出来了?这样不是很好啊......

遗憾,不能为这个赚分数。

就是上面的那个答案,这些题都比较简单,还是自己多动动脑筋吧!!!