\(x^3+\left(x-5\right)\left(x+8\right)=2x^3-37\)
\(x^3+x^2+3x-40=2x^3-37\)
\(\Leftrightarrow x^3-x^2-3x+3=0\)
\(\Leftrightarrow\left(x^2-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow x=1\) hoặc \(x=\pm\sqrt{3}\)
Vậy:....
\(x^3+\left(x-5\right)\left(x+8\right)=2x^3-37\)
\(\Leftrightarrow x^3+x^2+3x-40-2x^3+37=0\)
\(\Leftrightarrow-x^3+x^2+3x-3=0\)
\(\Leftrightarrow x^3-x^2-3x+3=0\)
\(\Leftrightarrow x^2\left(x-1\right)-3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-3\right)\left(x-1\right)=0\)
Do \(x^2-3=x^2-4+1=\left(x-2\right)\left(x+2\right)+1\ge1\) với \(\forall x\) nên suy ra: \(x-1=0\Leftrightarrow x=1\)
\(x^3+\left(x-5\right)\left(x+8\right)=2x^3-37\)
\(\Leftrightarrow x^3+x^2+3x-40-2x^3+37=0\)
\(\Leftrightarrow-x^3+x^2+3x-3=0\)
\(\Leftrightarrow x^2\left(1-x\right)+3\left(x-1\right)=0\)
\(\Leftrightarrow x^2\left(1-x\right)-3\left(1-x\right)=0\)
\(\Leftrightarrow\left(x^2-3\right)\left(1-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-3=0\\1-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=3\\x=1\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}\left[{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\\x=1\end{matrix}\right.\)
