Question

[(x+2)/(x+3)-(x+1)/(x+2)+(x+4)/(x+5)-(x+3)/(x+4)]/[(x的平方+7x+

Answer 1

(x+2)/(x+3)-(x+1)/(x+2)+(x+4)/(x+5)-(x+3)/(x+4)

=[1-1/(x+3)]-[1-1/(x+2)]+[1-1/(x+5)]-[1-1/(x+4)]

=1/(x+2)-1/(x+3)+1/(x+4)-1/(x+5)

=(x+3-x-2)/(x+2)(x+3)+(x+5-x-4)/(x+4)(x+5)

=1/(x^2+5x+6)+1/(x^2+9x+20)

=(2x^2+14x+26)/(x+2)(x+3)(x+4)(x+5)

=2(x^2+7x+13)/(x+2)(x+3)(x+4)(x+5)

x^2+8x+15=(x+3)(x+5)

所以原式=[2(x^2+7x+13)/(x+2)(x+3)(x+4)(x+5)]*[(x+3)(x+5)/(x^2+7x+13)]

=2/(x+2)(x+4)

=2/(x^2+6x+8)