x+[1/(x-1)]+[(x^2-3x+4)/(1-x^2)]

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/13 08:38:35

原式=x+1/(x-1)-(x²-3x+4)/(x²-1)
=x+(x+1)/(x+1)(x-1)-(x²-3x+4)/(x+1)(x-1)
=x+(x+1-x²+3x-4)/(x+1)(x-1)
=x-(x²-4x+3)/(x+1)(x-1)
=x-(x-1)(x-3)/(x+1)(x-1)
=x-(x-3)/(x+1)
=(x²+x-x+3)/(x+1)
=(x²+3)/(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-1)(X-2)(x-3)...(x-50)+x(x-2)(X-3)...(X-50)+...+x(x-1)(x-2) 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 1+x+x^2+x^3+.+x^99+x^100 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