a) Mình nghĩ bạn ghi đề thiếu.

Sửa \(\rightarrow x^2-12x+36=0\)

\(\Leftrightarrow\left(x-6\right)^2=0\)

\(\Leftrightarrow x-6=0\)

\(\Rightarrow x=6\)

S = {6}

b) \(x^2-5x=0\)

\(\Leftrightarrow x\left(x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

S = { 0;5}

c) \(x^2+5x+20=0\)

\(\Leftrightarrow x^2+4x+5x+20=0\)

\(\Leftrightarrow x\left(x+4\right)+5\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+4=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-5\end{matrix}\right.\)

S = { -5;-4}

d) \(x^3-8=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)=0\)

Ta có: \(x-2=0\)

\(\Rightarrow x=2\)

S = {2}

e) \(x^2+8x+15=0\)

\(\Leftrightarrow x^2+3x+5x+15=0\)

\(\Leftrightarrow x\left(x+3\right)+5\left(x+3\right)=0\)

\(\Rightarrow\left(x+3\right)\left(x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)

S = { -5; -3}

f) \(x^3+x^2+x+1=0\)

\(\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)

\(\Rightarrow\left(x^2+1\right)\left(x+1\right)=0\)

Vì \(x^2+1>0\) \(\Rightarrow x+1=0\)

\(\Rightarrow x=-1\)

S = {-1}