(x+1)(x+2)(x+3)(x+4)-120

来源：学生作业帮编辑：拍题作业网作业帮分类：数学作业时间：2024/05/18 18:35:07

(x+1)(x+2)(x+3)(x+4)-120
=[(x+4)(x+1)][(x+2)(x+3)]-120
=(x^2+5x+4)(x^2+5x+6)-120
=(x^2+5x)^2+10(x^2+5x)+24-120
=(x^2+5x)^2+10(x^2+5x)-96
=(x^2+5x+16)(x^2+5x-6)
=(x^2+5x+16)(x+6)(x-1)

x+2/x+1-x+3/x+2-x+4/x+3+x+5/x+4 x^5+x^4 = (x^3-x)(x^2+x+1)+x^2+x |X-1|+|X-2|+|X-3|+|X-4|+|X-5|+|X-6|+|X-7|+|X-8|+|X-9|+|X-10| x(2-1/x)+x/(x^2-2x)÷(3-x)/(x^2-4) x^4+x^3+x^2+x+1=0,x^2006+x^2005+x^2004+x^2003+x^2002 |x^2-3x-4|>x+1 (x+1-1/1-x)÷(x-x²/x-1) (x-4/x-x-6+ x+2/x-3)÷x+1/x-3 x-1)(X-2)(x-3)...(x-50)+x(x-2)(X-3)...(X-50)+...+x(x-1)(x-2) x^4-x^3-2x^2+3x-1 (x-1)^3(x+1)（x-2）/(x+4) （x+1)(x+2)(x+3)(x+4）+1 x^4+x^3+2x^2+x+1