Đáp án: $S = \bigg\{\dfrac{-\sqrt[]{13}+5}{2}, 2\bigg\}$

Giải thích các bước giải:

$ĐKXĐ : x ≥ -1,x \neq 3$

Ta có : $\dfrac{x-2}{\sqrt[]{x+1}+x} - \dfrac{x^2-5x+6}{2x-6} = 0 $

$⇔\dfrac{x-2}{\sqrt[]{x+1}+x} - \dfrac{x^2-2x-3x+6}{2.(x-3)}  = 0 $

$⇔\dfrac{x-2}{\sqrt[]{x+1}+x} - \dfrac{(x-2).(x-3)}{2.(x-3)}  = 0 $

$⇔\dfrac{x-2}{\sqrt[]{x+1}+x} - \dfrac{x-2}{2}  = 0 $

$⇔(x-2).\bigg(\dfrac{1}{\sqrt[]{x+1}+x} - \dfrac{1}{2}\bigg)  = 0 $

$⇔  \left[ \begin{array}{l}x-2=0\\\dfrac{1}{\sqrt[]{x+1}+x} - \dfrac{1}{2}=0\end{array} \right. $

$⇔  \left[ \begin{array}{l}x=2\text{(Thỏa mãn ĐKXĐ)}\\\dfrac{1}{\sqrt[]{x+1}+x} - \dfrac{1}{2}=0(1)\end{array} \right. $

Phương trình $(1)$ trở thành :

$\dfrac{2-\sqrt[]{x+1}-x}{2.(\sqrt[]{x+1}+x)} =0$

$⇒2-x = \sqrt[]{x+1}$ $ĐK : x ≤ 2$

$⇔(2-x)^2 = x+1$

$⇔4-4x+x^2 = x+1$

$⇔x^2-5x+3=0$

$⇔\bigg(x^2-2.x.\dfrac{5}{2} + \dfrac{25}{4}\bigg) - \dfrac{13}{4}=0$

$⇔\bigg(x-\dfrac{5}{2}\bigg)^2 = \dfrac{13}{4}$

$⇔ \left[ \begin{array}{l}x-\dfrac{5}{2} = \dfrac{\sqrt[]{13}}{2}\\x-\dfrac{5}{2} = -\dfrac{\sqrt[]{13}}{2}\end{array} \right. $

$⇔ \left[ \begin{array}{l}x= \dfrac{\sqrt[]{13}+5}{2}\text{(Không thỏa mãn ĐK)}\\x = \dfrac{-\sqrt[]{13}+5}{2}\text{(Thỏa mãn ĐK)}\end{array} \right. $

Vậy phương trình đã cho có tập nghiệm $S = \bigg\{\dfrac{-\sqrt[]{13}+5}{2}, 2\bigg\}$

ĐKXĐ:`x≥-1;x\ne3`

Xét `(x^2-5x+6)/(2x-6)`

`⇔(x^2-2x-3x+6)/(2x-6)`

`⇔[x(x-2)-3(x-2)]/(2x-6)`

`⇔[(x-2)(x-3)]/[2(x-3)]`

`⇔(x-2)/2` 

`⇒` Phương trình trở thành

`(x-2)/(\sqrt{x+1}+x)-(x-2)/2=0`

`⇔[2(x-2)-(x-2)(\sqrt{x+1}+x)]/[2(\sqrt{x+1}+x)]=0`

`⇔2x-4-(x.\sqrt{x+1}-2\sqrt{x+1}+x^2-2x)=0`

`⇔2x-4-x.\sqrt{x+1}+2\sqrt{x+1}-x^2+2x=0`

`⇔-x^2+4x-4-\sqrt{x+1}(x-2)=0`

`⇔x^2-4x+4+\sqrt{x+1}(x-2)=0`

`⇔(x-2)^2+\sqrt{x+1}(x-2)=0`

`⇔(x-2)(x-2+\sqrt{x+1})=0`

TH1:

`x-2=0`

`⇔x=2`

TH2:

`x-2+\sqrt{x+1}=0`

`⇔\sqrt{x+1}=2-x` `(`ĐKXĐ:`2-x≥0⇔x≤2)`

`⇔x+1=4-4x+x^2`

`⇔x^2-5x+3=0`

`Δ= 13`

`⇒`\(\left[ \begin{array}{l}x=\dfrac{5+\sqrt{13}}{2}(KTM)\\x=\dfrac{5-\sqrt{13}}{2}\end{array} \right.\)