\(\left(x+2\right)\left(x-3\right)\left(17x^2-17x+8\right)=\left(x+2\right)\left(x-3\right)\left(x^2-17x+33\right)\)
=>\(17x^2-17x+8=x^2-17x+33\)
<=> \(16x^2-25=0\)
<=>\(\left(4x-5\right)\left(4x+5\right)=0\)
=> \(4x-5=0=>x=\dfrac{5}{4}\)
hoặc \(4x+5=0=>x=\dfrac{-5}{4}\)
(x+2)(x−3)(17x2−17x+8)=(x+2)(x−3)(x2−17x+33)
\(\Leftrightarrow\)(x+2)(x−3)(17x2−17x+8) - (x+2)(x−3)(x2−17x+33) = 0
\(\Leftrightarrow\)(x+2)(x−3).[(17x2−17x+8)-(x2−17x+33)] = 0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}\text{x+2 = 0}\\\text{x−3 = 0}\\\text{(17x^2−17x+8)-(x^2−17x+33) = 0}\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-2\\x=3\\17x^2-17x+8-x^2+17x-33=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\\16x^2-25=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\\\left(4x-5\right)\left(4x+5\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\\4x-5=0\\4x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\\4x=5\\4x=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\\x=\dfrac{5}{4}\\x=\dfrac{-5}{4}\end{matrix}\right.\)
Vậy S = \(\left\{-2;\dfrac{-5}{4};\dfrac{5}{4};3\right\}\)
