Đáp án:
$2x^2+5x-2=0$
$⇔(\sqrt[]{2}x)^2 + 2 . \sqrt[]{2}x . \dfrac{5\sqrt[]{2}}{4} + (\dfrac{5\sqrt[]{2}}{4})^2 -\dfrac{41}{8} =0$
$⇔(\sqrt[]{2}x +\dfrac{5\sqrt[]{2}}{4})^2 - \dfrac{41}{8}=0$
$⇔(\sqrt[]{2}x +\dfrac{5\sqrt[]{2}}{4})^2 - (\sqrt[]{\dfrac{41}{8}})^2=0$( $\text{HĐT : $a^2-b^2=(a-b)(a+b)$}$)
$⇔(\sqrt[]{2}x +\dfrac{5\sqrt[]{2}}{4} - \sqrt[]{\dfrac{41}{8}}).(\sqrt[]{2}x +\dfrac{5\sqrt[]{2}}{4} +\sqrt[]{\dfrac{41}{8}})=0$
$⇔$\(\left[ \begin{array}{l}\sqrt[]{2}x +\dfrac{5\sqrt[]{2}}{4} - \sqrt[]{\dfrac{41}{8}}=0\\\sqrt[]{2}x +\dfrac{5\sqrt[]{2}}{4} + \sqrt[]{\dfrac{41}{8}}=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=\dfrac{-5+\sqrt[]{41}}{4}\\x=-\dfrac{5+\sqrt[]{41}}{4}\end{array} \right.\)
$\text{Vậy phương trình có tập hợp nghiệm S={$\dfrac{-5+\sqrt[]{41}}{4} ; -\dfrac{5+\sqrt[]{41}}{4}$} }$
