Bài 9:
a) Ta có: \(\left(x-1\right)\left(x^2+5x-2\right)-\left(x^3-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+5x-2\right)-\left(x-1\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+5x-2-x^2-x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{1;\dfrac{3}{4}\right\}\)
b) Ta có: \(x^2+\left(x+2\right)\left(11x-7\right)=4\)
\(\Leftrightarrow x^2+11x^2-7x+22x-14-4=0\)
\(\Leftrightarrow12x^2+15x-18=0\)
\(\Leftrightarrow3\left(4x^2+5x-6\right)=0\)
\(\Leftrightarrow4x^2+5x-6=0\)
\(\Leftrightarrow4x^2+8x-3x-6=0\)
\(\Leftrightarrow4x\left(x+2\right)-3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(4x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\4x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\4x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{-2;\dfrac{3}{4}\right\}\)
c) Ta có: \(x^3+1=x\left(x+1\right)\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1-x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
Vậy: S={-1;1}
d) Ta có: \(x^3+x^2+x+1=0\)
\(\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\cdot\left(x^2+1\right)=0\)
mà \(x^2+1>0\forall x\)
nên x+1=0
hay x=-1
Vậy: S={-1}
