Question

Giải pt sau:\(x^3-3x^2+9x-9=0\)

Answer 1

phương trình vô tỉ

Answer 2

dùng sơ đồ hooc ne nha bn ! 

Answer 3

\(x^3-3x^2+9x-9=0\)

\(x\left(x^2-3x+9\right)-9=0\)

\(x\left(x^2-3x+9\right)=0+9\)

\(x\left(x^2-3x+9\right)=9\)

\(\Rightarrow\text{ }x\text{ },\text{ }\left(x^2-3x+9\right)\inƯ\left(9\right)\)

Ta có bảng :

x	- 1	1	- 3	3	- 9	9
x2 - 3x + 9	- 9	9	- 3	3	- 1	1

                     \(\Rightarrow\text{ }x\in\left\{-1\text{ ; }-1\text{ ; }-3\text{ ; }3\text{ ; }-9\text{ ; }9\right\}\)

Answer 4

TL:

\(x^3-3x^2+9x-9=0\)

\(\Rightarrow x^2\left(x-3\right)+9\left(x-3\right)+18=0\)

\(\Rightarrow\left(x^2+9\right)\left(x-3\right)=-18\)

\(\Rightarrow x^2+9\inƯ\left(-18\right)\)

\(\Rightarrow x^2\in\left\{\pm1;\pm3;\pm2;\pm6;\pm9\right\}\)

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