Xét ptrinh
$\dfrac{1}{x(x+1)} + \dfrac{1}{(x+2)(x+3)} + \dfrac{1}{(x+3)(x+4)} + \cdots + \dfrac{1}{(x+9)(x+10)} = \dfrac{3x+1}{x(x+10)}$
$<-> \dfrac{1}{x(x+1)} + \dfrac{1}{x+2} - \dfrac{1}{x+3} + \dfrac{1}{x+3} - \dfrac{1}{x+4} + \cdots + \dfrac{1}{x+9} - \dfrac{1}{x+10} = \dfrac{3x+1}{x(x+10)}$
$<-> \dfrac{1}{x} - \dfrac{1}{x+1} + \dfrac{1}{x+2} - \dfrac{1}{x+10} = \dfrac{3x+1}{x(x+10)}$
$<-> (x+1)(x+2)(x+10) - x(x+2)(x+10) + x(x+1)(x+10) - x(x+1)(x+2) = (3x+1)(x+1)(x+2)$
$<-> (x^2 + 3x + 2)(x+10) - (x^2 +2x)(x+10) + (x^2 + x)(x+10) - (x^2 +x)(x+2) = (x^2 + 3x + 2)(3x+1)$
$<-> (x+10)(x^2 + 3x + 2 - x^2 - 2x + x^2 +x) - (x^3 +3x^2 + 2x) = (x^2 + 3x + 2)(3x+1)$
$<-> (x+10)(x^2 + 2x + 2) - (x^3 + 3x^2 + 2x) = 3x^3 +10x^2 +9x + 2$
$<-> x^3 +12x^2 +22x + 20 - x^3 - 3x^2 -2x = 3x^3 + 10x^2 + 9x + 2$'
$<-> 3x^3 +x^2 -11x -18 =0$
Ptrinh trên có duy nhất 1 nghiệm thực là $x = \dfrac{-5242+\sqrt{98516497}}{2005}$.
Vậy tập nghiệm $S = \left\{ \dfrac{-5242+\sqrt{98516497}}{2005} \right\}$.
