Đáp án:

`x\in {{-\sqrt{97}-13}/6;{\sqrt{97}-13}/6;`

`{\sqrt{17}+5}/2;{-\sqrt{17}+5}/2}`

Giải thích các bước giải:

Ta có:

`\qquad 3x^4-2x^3-53x^2-4x+12=0`

`<=>3x^4-15x^3+13x^3+6x^2+6x^2-65x^2+26x-30x+12=0`

`<=>(3x^4+13x^3+6x^2)-(15x^3+65x^2+30x)+(6x^2+26x+12)=0`

`<=>x^2(3x^2+13x+6)-5x(3x^2+13x+6)+2(3x^2+13x+6)=0`

`<=>(3x^2+13x+6)(x^2-5x+2)=0` 

`<=>[3(x^2+2x. {13}/6+({13}/6)^2)-{97}/{12}].\ [ (x^2-2x. 5/2+(5/2)^2)-{17}/4]=0`

`<=>[3(x+{13}/6)^2-{97}/{12}].\ [(x-5/2)^2-{17}/4]=0`

`<=>`$\left[\begin{array}{l}3(x+\dfrac{13}{6})^2-\dfrac{97}{12}=0\\(x-\dfrac{5}{2})^2-\dfrac{17}{4}=0\end{array}\right.$

`<=>`$\left[\begin{array}{l}(x+\dfrac{13}{6})^2=\dfrac{97}{36}\\(x-\dfrac{5}{2})^2=\dfrac{17}{4}\end{array}\right.$

`<=>`$\left[\begin{array}{l}x+\dfrac{13}{6}=\sqrt{\dfrac{97}{36}}=\dfrac{\sqrt{97}}{6}\\x+\dfrac{13}{6}=-\sqrt{\dfrac{97}{36}}=\dfrac{-\sqrt{97}}{6}\\x-\dfrac{5}{2}=\sqrt{\dfrac{17}{4}}=\dfrac{\sqrt{17}}{2}\\x-\dfrac{5}{2}=-\sqrt{\dfrac{17}{4}}=\dfrac{-\sqrt{17}}{2}\end{array}\right.$

`<=>`$\left[\begin{array}{l}x=\dfrac{\sqrt{97}-13}{6}\\x=\dfrac{-\sqrt{97}-13}{6}\\x=\dfrac{\sqrt{17}+5}{2}\\x=\dfrac{-\sqrt{17}+5}{2}\end{array}\right.$

Vậy các giá trị của `x` thỏa mãn đề bài là:

`x\in {{-\sqrt{97}-13}/6;{\sqrt{97}-13}/6;`

`\qquad {\sqrt{17}+5}/2;{-\sqrt{17}+5}/2}`