Question

1/x²+x+1/x²+3x+2+1/x²+5x+6+1/x²+7x+12,分式方程，急急急！

Answer 1

原式=1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)

=1/x-1/(x+4)
=4/(x²+4x)

Answer 2

1/x²+x+1/x²+3x+2+1/x²+5x+6+1/x²+7x+12
=(1/x²+1/x²+1/x²+1/x²）+(x+3x+5x+7x)+(2+6+12)
=4/x²+16x+20

追问

1/1-x+1/1+x+2/1+x^2+4/1+x ^4+8/1+x^8