a) Giả sử: (2x-3)² + 4x = 6
⇔ 4x² - 12x + 9 + 4x = 6
⇔ 4x² - 8x + 3 = 0
⇔ 4x² - 6x - 2x + 3 = 0
⇔ 2x (2x - 3) - (2x - 3) = 0
⇔ (2x - 1)(2x - 3) = 0
⇔ \(\left[ \begin{array}{l}2x - 1 = 0\\2x - 3 = 0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}2x = 1\\2x = 3\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x = $\frac{1}{2}$ \\x = $\frac{3}{2}$ \end{array} \right.\)
Vậy: nếu x ∈ { $\frac{1}{2}$ ; $\frac{3}{2}$} thì (2x - 3)² + 4x = 6
b) 2a² - 7a - 4 = 0
⇔ 2a² - 8a + a - 4 = 0
⇔ 2a(a - 4) + a - 4 = 0
⇔ (2a + 1)(a - 4) = 0
⇔ \(\left[ \begin{array}{l}2a + 1 = 0\\a - 4 = 0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}2a = -1\\a = 0 + 4\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}a = $\frac{-1}{2}$ \\a = 4 \end{array} \right.\)
Vậy: S={ $\frac{-1}{2}$ ; 4}
Cho mình xin câu trả lời hay nhất
