a)\(x^3+\left(-x^2+4x^2\right)+\left(-4x+5x\right)-5=\left(x^3-x^2\right)+\left(4x^2-4x\right)+\left(5x-5\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)+4x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(x^2+4x+5\right)=\left(x-1\right)\left[\left(x+2\right)^2+1\right]=0\)
\(\left[\begin{matrix}x-1=0\Rightarrow x=1\\\left(x+2\right)^2+1=0.Vo.N_o\end{matrix}\right.\) Vậy x=1 là nghiệm duy nhất
Có : \(x\left(x-1\right)\left(x+1\right)\left(x-2\right)=24\)
\(\Leftrightarrow\) \(\left(x^2-x\right)\left(x^2-x-2\right)=24\)
Đặt \(y=x^2-x\)
\(\Rightarrow\) \(y\left(y-2\right)=24\)
\(\Leftrightarrow\) \(y^2-2y-24=0\)
\(\Leftrightarrow\) \(\left(y+4\right)\left(y-6\right)=0\)
\(\Leftrightarrow\) \(\left[\begin{matrix}y=-4\\y=6\end{matrix}\right.\)
Với \(y=-4\) thì \(x^2-x=-4\)
\(\Rightarrow\) \(x^2-x+4=0\) vô nghiệm
Với \(y=6\) thì \(x^2-x=6\)
\(\Rightarrow\) \(x^2-x-6=0\)
\(\Leftrightarrow\) \(\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\) \(\left[\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Vậy \(S=\left\{-2;3\right\}\)
