Đáp án+Giải thích các bước giải:
$a)$$\dfrac{2x+1}{x-1}=$ $\dfrac{5(x-1)}{x+1}$
$⇔$$\dfrac{(2x+1)(x+1)}{(x+1)(x-1)}=$ $\dfrac{5(x-1)^2}{(x+1)(x-1)}$
$⇒2x^2+3x+1=5(x^2-2x+1)$
$⇔2x^2+3x+1=5x^2-10x+5$
$⇔5x^2-2x^2-10x-3x+5-1=0$
$⇔3x^2-13x+4=0$
$⇔(x-4)(3x-1)=0$
$⇔\left[ \begin{array}{l}x-4=0\\3x-1=0\end{array} \right. $
$⇔\left[ \begin{array}{l}x=4\\x=\dfrac{1}{3}\end{array} \right.$
$b)$$\dfrac{x-3}{x-2}+$ $\dfrac{x-2}{x-4}=-1$
$⇔$$\dfrac{(x-3)(x-4)}{(x-2)(x-4)}+$ $\dfrac{(x-2)^2}{(x-2)(x-4)}=-1$
$⇒x^2-7x+12+x^2-4x+4=-x^2+6x-8$
$⇔2x^2-11x+16+x^2-6x+8=0$
$⇔3x^2-17x+24=0$
$⇔(x-3)(3x-8)=0$
$⇔\left[ \begin{array}{l}x=3\\x=\dfrac{8}{3}\end{array} \right. $
$c)$$\dfrac{1}{x-1}+$ $\dfrac{2x^2-5}{x^3-1}=$ $\dfrac{4}{x^2+x+1}$
$⇔$$\dfrac{x^2+x+1}{x^3-1}+$ $\dfrac{2x^2-5}{x^3-1}=$ $\dfrac{4(x-1)}{x^3-1}$
$⇒ 3x^2+x-4=4x-4$
$⇔3x^2+x-4-4x+4=0$
$⇔3x^2-3x=0$
$⇔3x(x-1)=0$
$⇔$\(\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\)
$d)$$\dfrac{13}{(x-3)(2x+7)}+$ $\dfrac{1}{2x+7}=$ $\dfrac{6}{x^2-9}$
$⇔$$\dfrac{13(x+3)}{(x^2-9)(2x+7)}+$ $\dfrac{x^2-9}{(x^2-9)(2x+7)}=$ $\dfrac{6(2x+7)}{(x^2-9)(2x+7)}$
$⇒13x+39+x^2-9=12x+42$
$⇔x^2+13x+30-12x-42=0$
$⇔x^2+x-12=0$
$⇔(x-3)(x+4)=0$
$⇔\left[ \begin{array}{l}x=3\\x=-4\end{array} \right.$
