Đáp án:

Phương trình có tập nghiệm $S=\{-13;2\}.$

Giải thích các bước giải:

$\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\\ \text{ĐKXĐ: }  \left\{\begin{array}{l}x^2+9x+20\ne 0 \\ x^2+11x+30\ne 0 \\x^2+13x+42 \ne 0 \end{array} \right.\\ \dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\\ \Leftrightarrow \dfrac{1}{x^2+4x+5x+20}+\dfrac{1}{x^2+5x+6x+30}+\dfrac{1}{x^2+6x+7x+42}=\dfrac{1}{18}\\ \Leftrightarrow \dfrac{1}{x(x+4)+5(x+4)}+\dfrac{1}{x(x+5)+6(x+5)}+\dfrac{1}{x(x+6)+7(x+6)}=\dfrac{1}{18}\\ \Leftrightarrow \dfrac{1}{(x+4)(x+5)}+\dfrac{1}{(x+5)(x+6)}+\dfrac{1}{(x+6)(x+7)}=\dfrac{1}{18}\\ \Leftrightarrow \dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\\ \Leftrightarrow \dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\\ \Leftrightarrow \dfrac{x+7-x-4}{(x+4)(x+7)}=\dfrac{1}{18}\\ \Leftrightarrow \dfrac{3}{(x+4)(x+7)}=\dfrac{1}{18}\\ \Leftrightarrow (x+4)(x+7)=54\\ \Leftrightarrow x^2+11x+28=54\\ \Leftrightarrow x^2+11x-26=0\\ \Leftrightarrow x^2-2x+13x-26=0\\ \Leftrightarrow x(x-2)+13(x-2)=0\\ \Leftrightarrow (x+13)(x-2)=0\\ \Leftrightarrow x=-13;x=2$

Kiểm tra lại điều kiện ta thấy hai giá trị trên thoả mãn

Vậy phương trình có tập nghiệm $S=\{-13;2\}.$

\(x^2+9x+20=(x+4)(x+5)\\x^2+11x+30=(x+5)(x+6)\\x^2+13x+42=(x+6)(x+7)\\\text{ĐKXĐ:}x\ne -4,-5,-6,-7\\\dfrac{1}{x^2+9x+20}+\dfrac{1}{11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\\\leftrightarrow \dfrac{1}{(x+4)(x+5)}+\dfrac{1}{(x+5)(x+6)}+\dfrac{1}{(x+6)(x+7)}=\dfrac{1}{18}\\\leftrightarrow \dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\\\leftrightarrow \dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\\\leftrightarrow \dfrac{x+7-x-4}{x^2+11x+28}=\dfrac{1}{18}\\\leftrightarrow x^2+11x+28=54\\\leftrightarrow x^2+11x-26=0\\\leftrightarrow (x-2)(x+13)=0\\\leftrightarrow\left[ \begin{array}{l}x=2\\x=-13\end{array} \right.\text{(Tm)}\)

Phương trình có nghiệm \(x=2,x=-13\)