$ĐKXĐ:x \neq -4;-5;-6;-7$

$pt⇔\dfrac{1}{x^2+4x+5x+20}+\dfrac{1}{x^2+5x+6x+30}+\dfrac{1}{x^2+6x+7x+42}=\dfrac{1}{18}$

$⇔\dfrac{1}{(x+4)(x+5)}+\dfrac{1}{(x+5)(x+6)}+\dfrac{1}{(x+6)(x+7)}=\dfrac{1}{18}$

$⇔\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}$

$⇔\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}$

$⇔\dfrac{3}{(x+4)(x+7)}=\dfrac{1}{18}$

$⇔x^2+11x+28=54$

$⇔x^2+11x-26=0$

$⇔x^2-2x+13x-26=0$

$⇔(x-2)(x+13)=0$

$⇔$ \(\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)(t/m)

Vậy phương trình đã cho có tập nghiệm $S=(2;-13)$

Ta có: 

\(x^2+9x+2x=\left(x+4\right)\left(x+5\right)\)

\(x^2+11x+30=\left(x+6\right)\left(x+5\right)\)

\(x^2+13x+42=\left(x+6\right)\left(x+7\right)\)

ĐK: \(\left\{{}\begin{matrix}x\ne-4\\x\ne-5\\x\ne-6\\x\ne-7\end{matrix}\right.\)

pt \(\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{18\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}-\dfrac{18\left(x+4\right)}{18\left(x+4\right)\left(x+7\right)}=\dfrac{\left(x+4\right)\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}\)

\(\Rightarrow18\left(x+7\right)-18\left(x+4\right)=\left(x+4\right)\left(x+7\right)\)

\(\Leftrightarrow\left(x+13\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+13=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=2\end{matrix}\right.\)  (tm)

d: ĐKXĐ: x<>-4; x<>-5; x<>-6; x<>-7

\(PT\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

=>\(\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

=>\(\dfrac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)

=>x^2+11x+28=54

=>x^2+11x-26=0

=>(x+13)(x-2)=0

=>x=2 hoặc x=-13

e: \(\dfrac{x-241}{17}+\dfrac{x-220}{19}+\dfrac{x-195}{21}+\dfrac{x-166}{23}=10\)

\(\Leftrightarrow\left(\dfrac{x-241}{17}-1\right)+\left(\dfrac{x-220}{19}-2\right)+\left(\dfrac{x-195}{21}-3\right)+\left(\dfrac{x-166}{23}-4\right)=0\)

=>x-258=0

=>x=258

\(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\) (ĐKXĐ: \(x\notin\left\{-4;-5;-6;-7\right\}\))

<=> \(\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}-\dfrac{1}{18}=0\)

<=> \(\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}-\dfrac{1}{18}=0\)

<=> \(\dfrac{1}{x+4}-\dfrac{1}{x+7}-\dfrac{1}{18}=0\)

<=> \(\dfrac{18\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}-\dfrac{18\left(x+4\right)}{18\left(x+4\right)\left(x+7\right)}-\dfrac{\left(x+4\right)\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}=0\)

=> \(18\left(x+7\right)-18\left(x+4\right)-\left(x+4\right)\left(x+7\right)=0\)

<=> 18x + 18.7 - 18x - 18.4 - x² - 7x - 4x - 28 = 0

<=> - x² - 11x + 26 = 0

<=> (x - 2)(x + 13) = 0

<=> \(\left[{}\begin{matrix}x-2=0\\x+13=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\) (nhận)

Vậy S = {-13; 2}

ĐKXĐ: \(x\ne-4;-5;-6;-7\)

\(\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{3}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow x^2+11x-26=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)

\(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\) ĐKXĐ:x\(\ne\)-4,-5,-6,-7

\(\Leftrightarrow\)\(\dfrac{1}{x^2+4x+5x+20}+\dfrac{1}{x^2+5x+6x+30}+\dfrac{1}{x^2+6x+7x+42}=\dfrac{1}{18}\)

\(\Leftrightarrow\)\(\dfrac{1}{x.\left(x+4\right)+5.\left(x+4\right)}+\dfrac{1}{x.\left(x+5\right)+6.\left(x+5\right)}+\dfrac{1}{x.\left(x+6\right)+7.\left(x+6\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\)\(\dfrac{1}{\left(x+4\right).\left(x+5\right)}+\dfrac{1}{\left(x+5\right).\left(x+6\right)}+\dfrac{1}{\left(x+6\right).\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\)\(\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\)\(\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\)\(\dfrac{x+7-x-4}{\left(x+4\right).\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\)\(\dfrac{3}{\left(x+4\right).\left(x+7\right)}=\dfrac{3}{54}\)

\(\Leftrightarrow\)(x+4).(x+7)=54

\(\Leftrightarrow\)x²+11x+28=54

\(\Leftrightarrow\)x²+11x-26=0

\(\Leftrightarrow\)x²+13x-2x-26=0

\(\Leftrightarrow\)x.(x+13)-2.(x+13)=0

\(\Leftrightarrow\)(x-2).(x+13)=0

\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-2=0\\x+13=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2\left(TM\right)\\x=-13\left(TM\right)\end{matrix}\right.\)

Vậy tập nghiệm của pt trên là S={-13;2}

ĐKXĐ: \(x\ne-4;x\ne-5;x\ne-6;x\ne-7\)

\(\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{3}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow54=x^2+11x+28\)

\(\Leftrightarrow x^2+11x-26=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(TM\right)\\x=-13\left(TM\right)\end{matrix}\right.\)

\(pt\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)

ĐKXĐ: \(x\ne-4;-5;-6;-7\)

\(\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{3}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow54=x^2+11x+28\)

\(\Leftrightarrow x^2+11x-26=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)

ĐKXĐ: x khác -4; -5 ; -6 ; -7

\(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)\(\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{3}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow54=x^2+11x+28\)

\(\Leftrightarrow x^2+11x-26=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)

Vậy.........

\(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\left(đkxđ:x\ne-4;-5;-6;-7\right)\)

\(\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{3}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow x^2+11x+28=54\)

\(\Leftrightarrow x^2+11x-26=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-13\left(tm\right)\end{matrix}\right.\)