题目是（x+1）（x+2）（x+3）（x+4）+1怎么因式分解啊

来源：百度知道 编辑：UC知道 时间：2024/07/03 03:55:45

（x+1)(x+2)(x+3)(x+4)+1
=[(x+1)(x+4)][(x+2)(x+3)]+1
=(x²+5x+4)(x²+5x+6)+1
=[(x²+5x)+4][(x²+5x)+6]+1
=(x²+5x)²+10(x²+5x)+24+1
=(x²+5x)²+10(x²+5x)+25
=[(x²+5x)+5]²
=(x²+5x+5)².

解:
(x+1)(x+2)(x+3)(x+4)+1
=[(x+1)(x+4)][(x+2)(x+3)]+1
=(x^2+5x+4)(x^2+5x+6)+1
=(x^2+5x+4)[(x^2+5x+4)+2]+1
=(x^2+5x+4)^2+2(x^2+5x+4)+1
=(x^2+5x+4+1)^2
=(x^2+5x+5)^2

=[(x+1)(x+4)}{(x+2)(x+3)]+1
=[(x^2+5x)+4][(x^2+5x)+6]+1
=(x^2+5x)^2+10(x^2+5x)+24+1
=(x^2+5x)^2+10(x^2+5x)+25
=(x^2+5x+5)^2

[(x+1)(x+4)][(x+2)(x+3)]+1
=[x²+5x+4][x²+5x+6]+1
令x²+5x=A
=[A+4][A+6]+1
=A²+10A+25
=(A+5)²
=(x²+5x)²
=[x(x+5)]²
=x²(x+5)²
好了