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Trả lời:
ĐKXĐ: $x\neq 1$
$\dfrac{x^2}{(x-1)^2}=3x^2+4x$
$⇔x.\bigg{(}3x+4-\dfrac{x}{(x-1)^2}\bigg{)}=0$
$⇔\left[ \begin{array}{l}x=0\\3x+4-\dfrac{x}{(x-1)^2}=0\,\,(1)\end{array} \right.$
$(1)⇔3x-2+6-\dfrac{x}{x^2-2x+1}=0$
$⇔3x-2+\dfrac{6x^2-13x+6}{x^2-2x+1}=0$
$⇔3x-2+\dfrac{(3x-2)(2x-3)}{x^2-2x+1}=0$
$⇔(3x-2).\bigg{(}1+\dfrac{2x-3}{x^2-2x+1}\bigg{)}=0$
$⇔\left[ \begin{array}{l}x=\dfrac{2}{3}\\1+\dfrac{2x-3}{x^2-2x+1}=0\,\,(2)\end{array} \right.\,\,\,(2)⇔\dfrac{x^2-2}{x^2+2x-1}=0$
$⇔x^2=2⇔x=±\sqrt{2}$
Vậy `S={0;\frac{2}{3};\sqrt{2};-\sqrt{2}}`.
