x^3+2x^2-6x+1=2 căn x^2-x+1 -(x+1) căn 3x+1

Đáp án:

`S={0;1}` 

Giải thích các bước giải:

`x^3+2x^2-6x+1=2\sqrt{ x^2-x+1} -(x+1) \sqrt{ 3x+1}\ (1)`

`ĐKXĐ:`$\begin{cases}x^2-x+1\ge 0\\3x+1\ge 0\end{cases}$

`<=>`$\begin{cases}(x-\dfrac{1}{2})^2+\dfrac{3}{4}\ge 0\ (đúng\ \forall \ x\in R)\\x\ge \dfrac{-1}{3}\end{cases}$

`<=>x\ge -1/ 3`

`(1)<=>x^3+2x^2-6x+1-2\sqrt{x^2-x+1}+(x+1)\sqrt{3x+1}=0`

`<=>x^3-x+2(x^2-x+1)-2\sqrt{x^2-x+1}+(x+1)\sqrt{3x+1}-(3x+1)=0`

`<=>x^3-x^2+x^2-x+2\sqrt{x^2-x+1}.(\sqrt{x^2-x+1}-1)+\sqrt{3x+1}(x+1-\sqrt{3x+1})=0`

`<=>x(x^2-x)+x^2-x+{2\sqrt{x^2-x+1}.(\sqrt{x^2-x+1}-1)(\sqrt{x^2-x+1}+1)}/{\sqrt{x^2-x+1}+1}+{\sqrt{3x+1}(x+1-\sqrt{3x+1})(x+1+\sqrt{3x+1})}/{x+1+\sqrt{3x+1}}=0`

`<=>(x^2-x)(x+1)+{2\sqrt{x^2-x+1}.(x^2-x+1-1^2)}/{\sqrt{x^2-x+1}+1}+{\sqrt{3x+1}[(x+1)^2-(3x+1)]}/{x+1+\sqrt{3x+1}}=0`

`<=>(x^2-x)(x+1)+{2\sqrt{x^2-x+1}.(x^2-x)}/{\sqrt{x^2-x+1}+1}+{\sqrt{3x+1}.(x^2-x)}/{x+1+\sqrt{3x+1}}=0`

`<=>(x^2-x)(x+1+{2\sqrt{x^2-x+1}}/{\sqrt{x^2-x+1}+1}+\sqrt{3x+1}/{x+1+\sqrt{3x+1}})=0\ (2)`

`forall x\ge -1/ 3` ta có:

`x+1\ge -1/ 3+1= 2/ 3>0`

`{2\sqrt{x^2-x+1}}/{\sqrt{x^2-x+1}+1}>0`

`\sqrt{3x+1}/{x+1+\sqrt{3x+1}}\ge 0`

`=>x+1+{2\sqrt{x^2-x+1}}/{\sqrt{x^2-x+1}+1}+\sqrt{3x+1}/{x+1+\sqrt{3x+1}}>0`

Do đó `(2)<=>x^2-x=0`

`<=>`$\left[\begin{array}{l}x=0\\x=1\end{array}\right.$ (thỏa mãn)

Vậy phương trình có tập nghiệm `S={0;1}`