2) Ta có: \(19-\left(x-5\right)^3=x\left(3-x^2\right)-24\left(x-6\right)\)
\(\Leftrightarrow19-\left(x^3-15x^2+75x-125\right)=3x-x^3-24x+144\)
\(\Leftrightarrow19-x^3+15x^2-75x+125=-x^3-21x+144\)
\(\Leftrightarrow-x^3+15x^2-75x+144+x^3+21x-144=0\)
\(\Leftrightarrow15x^2-54x=0\)
\(\Leftrightarrow x\left(15x-54\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\15x-54=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\15x=54\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{18}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{0;\dfrac{18}{5}\right\}\)
3) Ta có: \(x\left(5-x\right)\left(x+5\right)-4x\left(x+5\right)=2x+1-\left(2x-1\right)^2\)
\(\Leftrightarrow x\left(5-x\right)\left(5+x\right)-4x\left(x+5\right)=2x+1-\left(4x^2-4x+1\right)\)
\(\Leftrightarrow x\left(25-x^2\right)-4x^2-20x=2x+1-4x^2+4x-1\)
\(\Leftrightarrow25x-x^3-4x^2-20x-2x-1+4x^2-4x+1=0\)
\(\Leftrightarrow-x^3-x=0\)
\(\Leftrightarrow x\left(x^2+1\right)=0\)
mà \(x^2+1>0\forall x\)
nên x=0
Vậy: S={0}
