`(x+3)(x^2-5x+8)=(x+3).x^2`
`<=>(x+3)(x^2-5x+8-x^2)=0`
`<=>(x+3)(8-5x)=0`
`<=>` \(\left[ \begin{array}{l}x+3=0\\8-5x=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac85\\x=-3\end{array} \right.\)
Vậy `S={-3,8/5}`
`(x+3)(x^2-5x+8)=(x+3).x^2`
`<=>(x+3)(x^2-5x+8-x^2)=0`
`<=>(x+3)(-5x+8)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\-5x+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{8}{5}\end{matrix}\right.\)
Vậy `S={-3;8/5}`.
\(\left(x+3\right)\left(x^2-5x+8\right)=\left(x+3\right).x^2\\ \Leftrightarrow\left(x+3\right)\left(x^2-5x+8\right)-x^2\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-5x+8-x^2\right)=0\\ \Leftrightarrow\left(x+3\right)\left(8-5x\right)=0\\\Leftrightarrow x+3=0 \)
hoặc \(8-5x=0\)
\(x+3=0\\ \Leftrightarrow x=-3\)
\(8-5x=0\\ \Leftrightarrow-5x=-8\\ \Leftrightarrow5x=8\\ \Leftrightarrow x=\dfrac{8}{5}\)
Vậy pt có tn là \(S=\left\{-3;\dfrac{8}{5}\right\}\)