Đáp án:

`S={(\sqrt{5}-3)/2,(-\sqrt{5}-3)/2,(\sqrt{5}+3)/2,(-\sqrt{5}+3)/2}`

Giải thích các bước giải:

 `x^3+1/x^3=6(x+1/x)(x ne 0)`

`<=>(x+1/x)(x^2-1+1/x^2)=6(x+1/x)`

`<=>(x+1/x)(x^2-1+1/x^2-6)=0`

`<=>((x^2+1)/x)(x^2+1/x^2-7)=0`

`(x^2+1)/x ne 0(AA x)`

`=>x^2+1/x^2-7=0`

`=>x^2+2+1/x^2-9=0`

`<=>(x+1/x)^2-3=0`

`<=>(x+1/x+3)(x+1/x-3)=0`

`+)x+1/x+3=0`

`<=>(x^2+3x+1)/x=0`

`<=>x^2+3x+1=0`

`<=>x^2+3x+9/4=5/4`

`<=>(x+3/2)^2=5/4`

`<=>x=(+-\sqrt{5}-3)/2`

`+)x+1/x-3=0`

`<=>(x^2-3x+1)/x=0`

`<=>x^2-3x+1=0`

`<=>x^2-3x+9/4=5/4`

`<=>(x-3/2)^2=5/4`

`<=>x=(+-\sqrt{5}+3)/2`

Vậy `S={(\sqrt{5}-3)/2,(-\sqrt{5}-3)/2,(\sqrt{5}+3)/2,(-\sqrt{5}+3)/2}`

Đáp án :

Phương trình có tập nghiệm `S={(\sqrt{5}+3)/2; -(\sqrt{5}+3)/2; (\sqrt{5}-3)/2; -(\sqrt{5}-3)/2}`

Giải thích các bước giải :

`+)Đkxđ : x \ne 0`
`x^3+1/x^3=6(x+1/x)`
`<=>(x+1/x)(x^2-x.(1)/x+1/x^2)-6(x+1/x)=0`
`<=>(x+1/x)(x^2-1+1/x^2-6)=0`
`<=>(x+1/x)(x^2+2+1/x^2-1-2-6)=0`
`<=>(x+1/x)[(x+1/x)^2-9]=0`
`+)x+1/x=x^2/x+1/x=(x^2+1)/x >= 1 => (x^2+1)/x > 0 => (x^2+1)/x \ne 0`
`=>(x+1/x)^2-9=0`
`<=>(x+1/x)^2=9`
`<=>(x+1/x)^2=(+-3)^2`
`<=>x+1/x=+-3`
`<=>`\(\left[ \begin{array}{l}x+\dfrac{1}{x}=3\\x+\dfrac{1}{x}=-3\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}\dfrac{x^2}{x}+\dfrac{1}{x}=\dfrac{3x}{x}\\\dfrac{x^2}{x}+\dfrac{1}{x}=\dfrac{-3x}{x}\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x^2+1=3x\\x^2+1=-3x\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x^2-3x+1=0\\x^2+3x+1=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x^2-2.x.\dfrac{3}{2}+({\dfrac{3}{2}})^2-\dfrac{9}{4}+\frac{4}{4}=0\\x^2+2.x.\dfrac{3}{2}+(\dfrac{3}{2})^2-\dfrac{9}{4}+\dfrac{4}{4}=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}(x-\dfrac{3}{2})^2-\dfrac{5}{4}=0\\(x+\dfrac{3}{2})^2-\dfrac{5}{4}=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}(x-\dfrac{3}{2})^2=\dfrac{5}{4}\\(x+\dfrac{3}{2})^2=\dfrac{5}{4}\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}(x-\dfrac{3}{2})^2=(±\dfrac{\sqrt{5}}{2})^2\\(x+\dfrac{3}{2})^2=(±\dfrac{\sqrt{5}}{2})^2\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x-\dfrac{3}{2}=±\dfrac{\sqrt{5}}{2}\\x+\dfrac{3}{2}=±\dfrac{\sqrt{5}}{2}\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}\left[ \begin{array}{l}x-\dfrac{3}{2}=\dfrac{\sqrt{5}}{2}\\x-\dfrac{3}{2}=-\dfrac{\sqrt{5}}{2}\end{array} \right.\\\left[ \begin{array}{l}x+\dfrac{3}{2}=\dfrac{\sqrt{5}}{2}\\x+\dfrac{3}{2}=-\dfrac{\sqrt{5}}{2}\end{array} \right.\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}\left[ \begin{array}{l}x=\dfrac{\sqrt{5}}{2}+\dfrac{3}{2}\\x=-\dfrac{\sqrt{5}}{2}+\dfrac{3}{2}\end{array} \right.\\\left[ \begin{array}{l}x=\dfrac{\sqrt{5}}{2}-\dfrac{3}{2}\\x=-\dfrac{\sqrt{5}}{2}-\dfrac{3}{2}\end{array} \right.\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}\left[ \begin{array}{l}x=\dfrac{\sqrt{5}+3}{2}  (tm)\\x=-\dfrac{\sqrt{5}+3}{2}  (tm)\end{array} \right.\\\left[ \begin{array}{l}x=\dfrac{\sqrt{5}-3}{2}  (tm)\\x=-\dfrac{\sqrt{5}-3}{2}  (tm)\end{array} \right.\end{array} \right.\)
Vậy : Phương trình có tập nghiệm `S={(\sqrt{5}+3)/2; -(\sqrt{5}+3)/2; (\sqrt{5}-3)/2; -(\sqrt{5}-3)/2}`