Đáp án:
$-2x^2 +3x+1=0$
$⇔2x^2-3x-1=0$
$⇔(\sqrt[]{2}x)^2 - 2. \sqrt[]{2}x . \dfrac{3\sqrt[]{2}}{4} +\dfrac{9}{8}-\dfrac{17}{8}=0$
$⇔(\sqrt[]{2}x -\dfrac{3\sqrt[]{2}}{4})^2 - (\sqrt[]{\dfrac{17}{8}})^2=0$
$⇔(\sqrt[]{2}x-\dfrac{3\sqrt[]{2}}{4} -\sqrt[]{\dfrac{17}{8}}).(\sqrt[]{2}x-\dfrac{3\sqrt[]{2}}{4} +\sqrt[]{\dfrac{17}{8}})=0$
$⇔$\(\left[ \begin{array}{l}\sqrt[]{2}x-\dfrac{3\sqrt[]{2}}{4} -\sqrt[]{\dfrac{17}{8}}=0\\\sqrt[]{2}x-\dfrac{3\sqrt[]{2}}{4} +\sqrt[]{\dfrac{17}{8}}=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=\dfrac{3+\sqrt[]{17}}{4}\\x=\dfrac{3-\sqrt[]{17}}{4}\end{array} \right.\)
$\text{Vậy phương trình có tập nghiệm S={$\dfrac{3\sqrt[]{17}}{4} ; \dfrac{3-\sqrt[]{17}}{4}$} }$
