a, x2 + 11x = 0
x(x + 11) = 0
\(\Rightarrow\left[\begin{matrix}x=0\\x+11=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=0\\x=-11\end{matrix}\right.\)
b, (x2 - 1)(x2 - 9) = 0
\(\Rightarrow\left[\begin{matrix}x^2-1=0\\x^2-9=0\end{matrix}\right.\)\(\Rightarrow\left[\begin{matrix}x^2=1\\x^2=9\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=1\\x=-1\\x=3\\x=-3\end{matrix}\right.\)
Vậy \(x\in\left\{1;-1;3;-3\right\}\)
c, ( |x + 1| - 5)(x2 - 9) = 0
\(\Rightarrow\left[\begin{matrix}\left|x+1\right|-5=0\\x^2-9=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}\left|x+1\right|=5\\x^2=9\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x+1=5\\x+1=-5\\x=3\\x=-3\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=4\\x=-6\\x=3\\x=-3\end{matrix}\right.\)
Vậy \(x\in\left\{4;-6;3;-3\right\}\)
d, \(3x-16⋮x+2\)
\(\Rightarrow3x+6-22⋮x+2\)
\(\Rightarrow3\left(x+2\right)-22⋮x+2\)
Vì \(3\left(x+2\right)⋮x+2\) nên để \(3\left(x+2\right)-22⋮x+2\) thì \(22⋮x+2\)
\(\Rightarrow x+2\inƯ\left(22\right)=\left\{\pm1;\pm2;\pm11;\pm22\right\}\)
| x + 2 | 1 | -1 | 2 | -2 | 11 | -11 | 22 | -22 |
| x | -1 | -3 | 0 | -4 | 9 | -13 | 20 | -24 |
Vậy x = {-1;-3;0;-4;9;-13;20;-24}