giai pt sau (x+3)2(3x+8)(3x+10)

(x+3)2(3x+8)(3x+10)=0 \( \left[{}\begin{matrix}\left(x+3\right)^2=0\\3x+8=0\\3x+10=0\end{matrix}\right.\) \(\left[{}\begin{matrix}x=-3\\x=\dfrac{-8}{3}\\x=\dfrac{-10}{3}\end{matrix}\right.\) S=\(\left\{-3;\dfrac{-8}{3};\dfrac{-10}{3}\right\}\)Câu trả lời: (x+3)2(3x+8)(3x+10)=0 \( \left[{}\begin{matrix}\left(x+3\right)^2=0\\3x+8=0\\3x+10=0\end{matrix}\right.\) \(\left[{}\begin{matrix}x=-3\\x=\dfrac{-8}{3}\\x=\dfrac{-10}{3}\end{matrix}\right.\) S=\(\left\{-3;\dfrac{-8}{3};\dfrac{-10}{3}\right\}\)