Giải pt :< $\frac{2}{(x-4)(x-2)}$ +$\frac{x+3}{x-4}$=$\frac{x+1}{2-x}$

` 2/((x-4)(x-2))  +(x+3)/(x-4)=(x+1)/(2-x)` ĐK : ` x \ne 2;\ x \ne 4`

`\to 2/((x-4)(x-2))  + ( (x+3)(x-2))/((x-4)(x-2)) = – ((x+1)(x-4))/((x-4)(x-2))`

`\to ( 2 + (x+3)(x-2) + (x+1)(x-4))/((x-4)(x-2)) =  0`

`\to 2 + (x^2 + x -6) + (x^2 – 3x -4) =0`

`\to 2x^2 – 2x -8 = 0`

`\to x^2 – x -4 = 0`

` \to x^2 – x +1/4 -17/4 = 0`

`\to (x-1/2)^2 =17/4`

`\to` \(\left[ \begin{array}{l}x – \dfrac{1}{2} = \dfrac{\sqrt{17}}{2}\\\\x – \dfrac{1}{2} = \dfrac{-\sqrt{17}}{2}\end{array} \right.\) 

`\to` \(\left[ \begin{array}{l}x  = \dfrac{\sqrt{17}+1}{2}\\\\x = \dfrac{-\sqrt{17}+1}{2}\end{array} \right.\) 

Vậy ` x \in { \frac{-\sqrt{17}+1}{2}; \frac{\sqrt{17}+1}{2}}`