Ta có:

 \(\begin{array}{l} {x^2} + 3x + 2 = \left( {x + 1} \right)\left( {x + 2} \right)\\ {x^2} + 5x + 6 = \left( {x + 2} \right)\left( {x + 3} \right)\\ {x^2} + 7x + 12 = \left( {x + 3} \right)\left( {x + 4} \right)\\ {x^2} + 9x + 20 = \left( {x + 4} \right)\left( {x + 5} \right) \end{array}\)

Khi đó:

\( pt \Leftrightarrow \frac{1}{{\left( {x + 1} \right)\left( {x + 2} \right)}} + \frac{1}{{\left( {x + 2} \right)\left( {x + 3} \right)}} + \frac{1}{{\left( {x + 3} \right)\left( {x + 4} \right)}} + \frac{1}{{\left( {x + 4} \right)\left( {x + 5} \right)}} = \frac{1}{3}\)

ĐKXĐ: \(x≠−1;−2;−3;−4;−5\)

Khi đó: \(\begin{array}{l} pt \Leftrightarrow \frac{1}{{\left( {x + 1} \right)\left( {x + 2} \right)}} + \frac{1}{{\left( {x + 2} \right)\left( {x + 3} \right)}} + \frac{1}{{\left( {x + 3} \right)\left( {x + 4} \right)}} + \frac{1}{{\left( {x + 4} \right)\left( {x + 5} \right)}} = \frac{1}{3}\\ \Leftrightarrow \frac{{1\left( {x + 5} \right) - 1\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x + 5} \right)}} = \frac{{\left( {x + 1} \right)\left( {x + 5} \right)}}{{3\left( {x + 1} \right)\left( {x + 5} \right)}}\\ \Rightarrow 3\left[ {x + 5 - \left( {x + 1} \right)} \right] = \left( {x + 1} \right)\left( {x + 5} \right)\\ \Leftrightarrow 3\left( {x + 5 - x - 1} \right) = {x^2} + 6x + 5 \Leftrightarrow {x^2} + 6x - 7 = 0\\ \Leftrightarrow \left( {x - 1} \right)\left( {x + 7} \right) = 0 \to \left[ \begin{array}{l} x = 1\\ x = - 7 \end{array} \right. \end{array}\)

⇒S={1;−7} nên tổng bình phương các nghiệm là:

\(1^2+(−7)^2=50\)

Đáp án cần chọn là: D