Đáp án:
$\dfrac{x^2-5x+4}{x-1} + \dfrac{x^2-8x+4}{2x+1} = 0$
$\text{ĐKXĐ : x $\neq$ 1 ; x $\neq$ -$\dfrac{1}{2}$}$
$⇔\dfrac{(x^2-5x+4)(2x+1)}{(x-1)(2x+1)} + \dfrac{(x^2-8x+4)}{(x-1)(2x+1)} = 0$
$⇒ (x^2-5x+4)(2x+1) + (x^2-8x+4)(x-1) =0$
$⇔2x³+x² -10x² -5x +8x +4 + x³ -x² -8x² +8x +4x -4 =0$
$⇔3x³-18x²+15x=0$
$⇔3x³ -3x² -15x² +15x =0$
$⇔ 3x²(x-1)-15x(x-1) =0$
$⇔(x-1)(3x²-15x)= 0$
$⇔(x-1)(x-5)3x=0$
$⇔3x =0 ⇔x = 0 $ $\text{(TM)}$
⇔\(\left[ \begin{array}{l}x-1=0\\x-5=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=1(KTM)\\x=5(TM)\end{array} \right.\)
$\text{Vậy phương trình có tập nghiệm S={0 ; 5 }}$
