\(x^3-3x^2+9x-9=0\)
\(x\left(x^2-3x+9\right)-9=0\)
\(x\left(x^2-3x+9\right)=0+9\)
\(x\left(x^2-3x+9\right)=9\)
\(\Rightarrow\text{ }x\text{ },\text{ }\left(x^2-3x+9\right)\inƯ\left(9\right)\)
Ta có bảng :
| x | - 1 | 1 | - 3 | 3 | - 9 | 9 |
| x2 - 3x + 9 | - 9 | 9 | - 3 | 3 | - 1 | 1 |
\(\Rightarrow\text{ }x\in\left\{-1\text{ ; }-1\text{ ; }-3\text{ ; }3\text{ ; }-9\text{ ; }9\right\}\)
TL:
\(x^3-3x^2+9x-9=0\)
\(\Rightarrow x^2\left(x-3\right)+9\left(x-3\right)+18=0\)
\(\Rightarrow\left(x^2+9\right)\left(x-3\right)=-18\)
\(\Rightarrow x^2+9\inƯ\left(-18\right)\)
\(\Rightarrow x^2\in\left\{\pm1;\pm3;\pm2;\pm6;\pm9\right\}\)
Làm nốt
