Question

两道因式分解

n²+n²(n+1)²+(n+1)²
2a^4+2b^4+2c^4-2a²b²-2b²c²-2a²c²

Answer 1

解：n²+n²(n+1)²+(n+1)²=[n(n+1)]²+2n²+2n+1=[n(n+1)]²+2n(n+1)+1=[n(n+1)+1]²=(n²+n+1)²
2a^4+2b^4+2c^4-2a²b²-2b²c²-2a²c²=(a²-b²)²+(b²-c²)²+(a²-c²)²

追问

第二道不能再分解了吗？

追答

貌似不能分解。主元法，把a²当作未知数，原式=2(a²)²-2(b²+c²)a²+[2(b²)²+2(c²)²-2b²c²]
当b²≠c²时，判别式小于0.

Answer 2

（1）n^2+n^2*(n+1)^2+(n+1)^2
=n^2+n^2(n+1)*(n+1)+(n+1)^2
=n^2+n^3(n+1)+n^2(n+1)+(n+1)^2
=n^4+n^2(n+1)+n^2(n+1)+(n+1)^2
=(n^2+n+1)^2

Answer 3

（1）n^2+n^2*(n+1)^2+(n+1)^2
=n^2+n^2(n+1)*(n+1)+(n+1)^2
=n^2+n^3(n+1)+n^2(n+1)+(n+1)^2
=n^4+n^2(n+1)+n^2(n+1)+(n+1)^2
=(n^2+n+1)^2
2a^4+2b^4+2c^4-2a²b²-2b²c²-2a²c
=(a²-b²)²+(b²-c²)²+(a²-c²)²望采纳o(∩_∩)o...