đặt x+29=y
ta có :
\(\dfrac{1}{y^2}+\dfrac{1}{\left(y+1\right)^2}=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{2y^2+2y+1}{y^4+2y^3+y^2}=\dfrac{5}{4}\)
\(\Leftrightarrow5y^4+10y^3-3y^2-8y-4=0\)
\(\Leftrightarrow\left(y-1\right)\left(5y^3+15y^2+12y+4\right)=0\)
\(\Leftrightarrow\left(y-1\right)\left(y+2\right)\left(5y^2+5y+2\right)=0\)
\(\left[{}\begin{matrix}y-1=0\Leftrightarrow y=1\Rightarrow x+29=1\Leftrightarrow x=-28\\y+2=0\Leftrightarrow y=-2\Rightarrow x+29=-2\Leftrightarrow x=-31\\5y^2+5y+2=0\left(loại\right)\end{matrix}\right.\)
vậy, S={-31;-28}