(x - 3)/(x - 2) + (x - 2)/(x - 4) = -1 giải pt

$\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1$ $(ĐKXĐ:x≠2;x≠4)$

$⇔ \dfrac{(x-4)(x-3)}{(x-2)(x-4)}+\dfrac{(x-2)(x-2)}{(x-2)(x-4)}=\dfrac{-1(x-2)(x-4)}{(x-2)(x-4)}$

$⇔ \dfrac{x^2-7x+12}{(x-2)(x-4)}+\dfrac{x^2-4x+4}{(x-2)(x-4)}=\dfrac{-(x^2-6x+8)}{(x-2)(x-4)}$

$⇔ \dfrac{x^2-7x+12}{(x-2)(x-4)}+\dfrac{x^2-4x+4}{(x-2)(x-4)}=\dfrac{-x^2+6x-8}{(x-2)(x-4)}$

$⇒ x^2-7x+12+x^2-4x+4=-x^2+6x-8$

$⇔ 2x^2-11x+16=-x^2+6x-8$

$⇔ 2x^2+x^2-11x-6x+16+8=0$

$⇔ 3x^2-17x+24=0$

$⇔ 3x^2-8x-9x+24=0$

$⇔ x(3x-8)-3(3x-8)=0$

$⇔ (x-3)(3x-8)=0$

$⇔$ \(\left[ \begin{array}{l}x-3=0\\3x-8=0\end{array} \right.\) 

$⇔$ \(\left[ \begin{array}{l}x=3\\3x=8\end{array} \right.\) 

$⇔$ \(\left[ \begin{array}{l}x=3\\x=\dfrac{8}{3}\end{array} \right.\) (thỏa mãn ĐKXĐ)

Vậy phương trình có tập nghiệm `S={3;(8)/(3)}`

$#thanhmaii2008$