Đáp án:
$\dfrac{1}{x-1} +\dfrac{2}{x-2} +\dfrac{3}{x-3}=\dfrac{6}{x-6}$
$\text{ĐKXĐ : $x \neq 1 ; x \neq 2 ; x \neq 3 ; x \neq 6$}$
$⇔(x-6)(x-2)(x-3) +2(x-1)(x-3)(x-6) +3(x-1)(x-2)(x-6) = 6(x-1)(x-2)(x-3)$
$⇔x^3 -11x^2+36x -36 +2x^3 -20x^2+54x -36 +3x^3 -27x^2 +60x -36 =6x^3 -36x^2+66x-36$
$⇔x^3 +2x^3 +3x^3 -6x^3 -11x^2 -20x^2 -27x^2 +36x^2 +36x +54x +60x -66x -36 -36-36+36=0$
$⇔-22x^2 +84x -72=0$
$⇔22x^2 -84x +72=0$
$⇔[(\sqrt[]{22}x)^2- 2. \sqrt[]{22}x . \dfrac{21\sqrt[]{22}}{11}+ (\dfrac{21\sqrt[]{22}}{11})^2] -\dfrac{90}{11}=0$
$⇔(\sqrt[]{22}x -\dfrac{21\sqrt[]{22}}{11})^2 -(\sqrt[]{\dfrac{90}{11}})^2 =0$
$⇔(\sqrt[]{22}x -\dfrac{21\sqrt[]{22}}{11} -\sqrt[]{\dfrac{90}{11}}).(\sqrt[]{22}x -\dfrac{21\sqrt[]{22}}{11}+\sqrt[]{\dfrac{90}{11}})=0$
$⇔$\(\left[ \begin{array}{l}\sqrt[]{22}x-\dfrac{21\sqrt[]{22}}{11}-\sqrt[]{\dfrac{90}{11}}=0\\\sqrt[]{22}x-\dfrac{21\sqrt[]{22}}{11}+\sqrt[]{\dfrac{90}{11}}=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=\dfrac{21+3\sqrt[]{5}}{11}(chọn)\\x=\dfrac{21-3\sqrt[]{5}}{11}(chọn)\end{array} \right.\)
$\text{Vậy phương trình có tập nghiệm S={$\dfrac{21+3\sqrt[]{5}}{11} ; \dfrac{21-3\sqrt[]{5}}{11}$}}$
