1.(x*2-3x+3)/(x*2-3x+2)-(x*2-5x+7)/(x*2-5x+6)-1/(x*2-4x+3)
问题描述：
1.(x*2-3x+3)/(x*2-3x+2)-(x*2-5x+7)/(x*2-5x+6)-1/(x*2-4x+3)
2.1/x(x+1)+2/(x+1)(x+3)+3/(x+3)(x+6)3.已知：x*2-3x+1=0,求x*2+1/x*2.
答
1、
(x^2-3x+3)/(x^2-3x+2)-(x^2-5x+7)/(x^2-5x+6)-1/(x^2-4x+3)
=1+1/(x^2-3x+2) -1 -1/(x^2-5x+6)-1/(x^2-4x+3)
=1/(x^2-3x+2) -1/(x^2-5x+6)-1/(x^2-4x+3)
=1/(x-2) -1/(x-1) -1/(x-3) +1/(x-2) - 0.5*[1/(x-3) -1/(x-1)]
=2/(x-2) - 1/(2x-2) -3/(2x-6)
2、
1/x(x+1)+2/(x+1)(x+3)+3/(x+3)(x+6)
=1/x -1/(x+1) +1/(x+1) -1/(x+3) +1/(x+3) -1/(x+6)
=1/x -1/(x-6)
3、
x^2-3x+1=0
即x^2+1=3x,那么x+1/x=3,
等式两边平方得到
x^2+ 1/x^2 +2=9
即x^2+ 1/x^2=7第一道做错了。但，谢谢！