\(x^2\left(x+4,5\right)=13,5\)
<=>\(x^3+4,5x^2-13,5=0\)
<=> \(x^3+3x^2+1,5x^2+4,5x-4,5x-13,5=0\)
<=>\(x^2\left(x+3\right)+1,5x\left(x+3\right)-4,5\left(x+3\right)=0\)
<=>\(\left(x+3\right)\left(x^2+1,5x-4,5\right)=0\)
<=>\(\left(x+3\right)\left[x^2+3x-1,5-4,5\right]=0\)
<=>\(\left(x+3\right)\left[x\left(x+3\right)-1,5\left(x+3\right)\right]=0\)
<=>\(\left(x+3\right)^2\left(x-1,5\right)=0\)
<=> \(\left[{}\begin{matrix}\left(x+3\right)^2=0\\x-1,5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1,5\end{matrix}\right.\)
Vậy...
Ta có: \(x^2\left(x+4.5\right)=13.5\)
\(\Leftrightarrow x^3+\dfrac{9}{2}x^2-\dfrac{27}{2}=0\)
\(\Leftrightarrow2x^3+9x^2-27=0\)
\(\Leftrightarrow2x^3-3x^2+12x^2-18x+18x-27=0\)
\(\Leftrightarrow x^2\left(2x-3\right)+12x\left(2x-3\right)+9\left(2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x^2+12x+9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x^2+12x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\\left(x+6\right)^2=27\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x+6=3\sqrt{3}\\x+6=-3\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\sqrt{3}-6\\x=-3\sqrt{3}-6\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{3}{2};3\sqrt{3}-6;-3\sqrt{3}-6\right\}\)
