giải phương trình : a) 2 – x+1/x+2 = x+19/(x-3)(x-2) + 2/x-3 b) (2x+3)(x-5)=4x^2 +6x c) |3-x^2 | + (x-1)(x+5)-x^2+2

0 bình luận về “giải phương trình : a) 2 – x+1/x+2 = x+19/(x-3)(x-2) + 2/x-3 b) (2x+3)(x-5)=4x^2 +6x c) |3-x^2 | + (x-1)(x+5)-x^2+2”

1.  a) $2-\dfrac{x+1}{x-2} = \dfrac{x+19}{(x-3).(x-2)} + \dfrac{2}{x-3}$ $(1)$

   $ĐKXĐ : x \neq 2,x \neq 3$

   Phương trình $(1)$ tương đương :

   $\dfrac{2.(x-3).(x-2)}{(x-3).(x-2)}-\dfrac{(x+1).(x-3)}{(x-2).(x-3)} = \dfrac{x+19}{(x-3).(x-2)} + \dfrac{2.(x-2)}{(x-2).(x-3)}$

   $⇒2.(x-3).(x-2) -(x+1).(x-3) = x+19+2.(x-2)$

   $⇔(x-3).(x-5) = 3x +15$

   $⇔x^2+15-8x = 3x+15$

   $⇔x^2-11x = 0 $

   $⇔\left[ \begin{array}{l}x=0\\x=11\end{array} \right.$ ( Thỏa mãn )

   Vậy $S =\big\{0,11\big\}$

   b) $(2x+3).(x-5) = 4x^2+6x$

   $⇔(2x+3).(x-5) = 2x.(2x+3)$

   $⇔(2x+3).(x-5-2x)=0$

   $⇔(2x+3).(-x-5)=0$

   $⇔(2x+3).(x+5)=0$

   $⇔\left[ \begin{array}{l}2x+3=0\\x+5=0\end{array} \right.$

   $⇔\left[ \begin{array}{l}x=\dfrac{-3}{2}\\x=-5\end{array} \right.$

   Vậy $S= \bigg\{-\dfrac{3}{2}, -5\bigg\}$

   Bình luận