Đáp án + giải thích các bước giải:

$ĐKXD:\begin{cases} x^2-4x+3 \ge 0 (1)  \\2x^2-3x+1 \ge 0(2)  \\x-1\ge0(3) \end{cases} $

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

`->(x-1)(x-3)>=0`

`->`\(\left[ \begin{array}{l}x\ge3\\x\le1\end{array} \right.\) 

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

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

`->(2x-1)(x-1)>=0`

`->`\(\left[ \begin{array}{l}x\ge1\\x\le\dfrac{1}{2}\end{array} \right.\) 

`(3)->x>=1`

`\sqrt{x^2-4x+3}-\sqrt{2x^2-3x+1}=x-1`

`->((\sqrt{x^2-4x+3}-\sqrt{2x^2-3x+1})(\sqrt{x^2-4x+3}+\sqrt{2x^2-3x+1}))/(\sqrt{x^2-4x+3}+\sqrt{2x^2-3x+1})=x-1`

`->(x^2-4x+3-(2x^2-3x+1))/(\sqrt{x^2-4x+3}+\sqrt{2x^2-3x+1})=x-1`

`->(-x^2-x+2)/(\sqrt{x^2-4x+3}+\sqrt{2x^2-3x+1})=x-1`

`->-(x^2-x+2x-2)/(\sqrt{x^2-4x+3}+\sqrt{2x^2-3x+1})=x-1`

`->-((x-1)(x+2))/(\sqrt{x^2-4x+3}+\sqrt{2x^2-3x+1})-(x-1)=0`

`->((x-1)(x+2))/(\sqrt{x^2-4x+3}+\sqrt{2x^2-3x+1})+(x-1)=0`

`->(x-1)((x+2)/(\sqrt{x^2-4x+3}+\sqrt{2x^2-3x+1})+1)=0`

`->x-1=0`

`->x=1(TM)`

Vậy `S={1}`

Đáp án:

$S=\{1\}$

Giải thích các bước giải:

ĐKXĐ: $\begin{cases}x^2-4x+3\ge 0\\2x^2-3x+1\ge 0\end{cases}⇔\begin{cases}(x-1)(x-3)\ge 0\\(x-1)(2x-1)\ge 0\end{cases}$

$⇔\begin{cases}\left[ \begin{array}{l}x\le 1\\x\ge 3\end{array} \right.\\\left[ \begin{array}{l}x\le \dfrac{1}{2}\\x\ge 1\end{array} \right.\end{cases}⇔\left[ \begin{array}{l}x\le \dfrac{1}{2}\\x=1\\x\ge 3\end{array} \right.$

$\sqrt{x^2-4x+3}-\sqrt{2x^2-3x+1}=x-1$

$⇔\sqrt{x^2-4x+3}+1-x=\sqrt{2x^2-3x+1}$

Bình phương $2$ vế với điều kiện:

$\sqrt{x^2-4x+3}+1-x\ge 0$

$⇒\sqrt{x^2-4x+3}\ge x-1\,\,(ĐK:\,x\ge 1)$

$⇒x^2-4x+3\ge x^2-2x-1$

$⇒x\le 1$

$\text{Pt}⇔(\sqrt{x^2-4x+3}+1-x)^2=(\sqrt{2x^2-3x+1})^2$

$⇔x^2-4x+3+2\sqrt{x^2-4x+3}.(1-x)+(1-x)^2=2x^2-3x+1$

$⇔x^2-4x+3+1-2x+x^2+2\sqrt{x^2-4x+3}.(1-x)=2x^2-3x+1$

$⇔2\sqrt{x^2-4x+3}.(1-x)=3x-3$

$⇔3(x-1)+2\sqrt{x^2-4x+3}.(x-1)=0$

$⇔(x-1)(3+2\sqrt{x^2-4x+3})=0$

$⇔\left[ \begin{array}{l}x-1=0\\3+2\sqrt{x^2-4x+3}=0\end{array} \right.⇔\left[ \begin{array}{l}x=1\,(TM)\\2\sqrt{x^2-4x+3}=-3\,(VN)\end{array} \right.$

Vậy $S=\{1\}$.