Đáp án:
`S = {`$\dfrac{-4 +\sqrt{30}}{2}; \dfrac{-4 -\sqrt{30}}{2}$`}`
Giải thích các bước giải:
`Cách` `1:`
$2x² +8x -7 = 0$ (Áp dụng HĐT: (a ±b)² = a² ±2ab +b²)
$⇔ (\sqrt{2}x)² +2.\sqrt{2}x.2\sqrt{2} +(2\sqrt{2})² -(2\sqrt{2})² -7 = 0$
$⇔ (\sqrt{2}x +2\sqrt{2})² -15 = 0$
$⇔ (\sqrt{2}x +2\sqrt{2} -\sqrt{15}).(\sqrt{2}x +2\sqrt{2} +\sqrt{15}) = 0$
$⇔ \left[ \begin{array}{l}\sqrt{2}x +2\sqrt{2} -\sqrt{15}=0\\\sqrt{2}x +2\sqrt{2} +\sqrt{15}=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=\dfrac{-4 +\sqrt{30}}{2}\\x=\dfrac{-4 -\sqrt{30}}{2}\end{array} \right.$
Vậy `S = {`$\dfrac{-4 +\sqrt{30}}{2}; \dfrac{-4 -\sqrt{30}}{2}$`}`
`Cách` `2:`
$2x² +8x -7 = 0$
`⇔ 2.(x² +4x -7/2) = 0`
`⇔ 2.[(x)² +2.x.2 +(2)² -(2)² -7/2] = 0`
`⇔ 2.[(x +2)² -15/2] = 0`
$⇔ 2.(x +2 -\sqrt{\dfrac{15}{2}}).(x +2 +\sqrt{\dfrac{15}{2}}) = 0$
$⇔ \left[ \begin{array}{l}x +2 -\sqrt{\dfrac{15}{2}}=0\\x +2 +\sqrt{\dfrac{15}{2}}=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=\dfrac{-4 +\sqrt{30}}{2}\\x=\dfrac{-4 -\sqrt{30}}{2}\end{array} \right.$
Vậy `S = {`$\dfrac{-4 +\sqrt{30}}{2}; \dfrac{-4 -\sqrt{30}}{2}$`}`
