1/x2-3x+2+1/x2-5x+6+1/x2-7+1/x2-7x+12-4/x2-6x+5 有多少种写多少种解法,

=1/(x-2)(x-1)+1/(x-2)(x-3)+1/(x-3)(x-4)-4/(x-1)(x-5) 帮你删除了 1/x2-7
=1/(x-2)-1/(x-1)+1/(x-3)-1/(x-2)+1/(x-4)-1/(x-3)-[1/(x-5)-1/(x-1)]
=1/(x-5)