\(\left(x+2\right)\left(x+3\right)\left(17x^2-17x+8\right)=\left(x+2\right)\left(x+3\right)\left(x^2-17x+33\right)\\ \Leftrightarrow\left(x+2\right)\left(x+3\right)\left(17x^2-17x+8\right)-\left(x+2\right)\left(x+3\right)\left(x^2-17x+33\right)=0\\\Leftrightarrow \left(x+2\right)\left(x+3\right)\left(16x^2-25\right)=0\\\Leftrightarrow \left(x+2\right)\left(x+3\right)\left(4x-5\right)\left(4x+5\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x+2=0\\x+3=0\\4x-5=0\\4x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-3\\x=\frac{5}{4}\\x=-\frac{5}{4}\end{matrix}\right.\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{-2;-3;-\frac{5}{4};\frac{5}{4}\right\}\)
