Tìm x, biết: a) x^2 + 4 = 4x; b) 2x^2 + 7x + 3 = 0.
Tìm x, biết:
a) x2 + 4 = 4x;
b) 2x2 + 7x + 3 = 0.
a) x2 + 4 = 4x
\( \Leftrightarrow \)x2 + 4 – 4x = 0
\( \Leftrightarrow \)x2 – 2.x.2 + 22 = 0
\( \Leftrightarrow \)(x – 2)2 = 0
\( \Leftrightarrow \)x – 2 = 0
\( \Leftrightarrow \)x = 2.
Vậy x = 2.
b) 2x2 + 7x + 3 = 0
\( \Leftrightarrow \)2x2 + x + 6x + 3 = 0
\( \Leftrightarrow \)x(2x + 1) + 3(2x + 1) = 0
\( \Leftrightarrow \)(2x + 1)(x + 3) = 0
\( \Leftrightarrow \)\(\left[ \begin{array}{l}2x + 1 = 0\\x + 3 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{{ - 1}}{2}\\x = - 3\end{array} \right.\)
Vậy \(x = \left\{ { - \frac{1}{2}; - 3} \right\}\).