x(x+1)分之ㄧ+(x+1)(x+3)分之二+(x+3)(x+6)分之三=x(x+6)分之k,試求常數k的值?

1/(x(x+1)) + 2/((x+1)(x+3)) + 3/((x+3)(x+6))
= (x+1 - x)/(x(x+1)) + (x+3 - (x+1))/((x+1)(x+3)) + (x+6 - (x+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)
= 6/(x(x+6))
常數 k = 6

1/[x(x+1)]+2/[(x+1)(x+3)]+3/[(x+3)(x+6)=k/[x(x+6)]
left=[(x+3)(x+6)+2x(x+6)+3x(x+1)]/[x(x+1)(x+3)(x+6)
left=(x^2+9x+18+2x^2+12x+3x^3+3x)/[x(x+1)(x+3)(x+6)]
left=(6x^2+24x+18)/[x(x+1)(x+3)(x+6)]
left=6(x^2+4x+3)/[x(x+1)(x+3)(x+6)]
left=6(x+1)(x+3)/[x(x+1)(x+3)(x+6)]
left=6/[x(x+6)
k=6