Đáp án:
\(\begin{array}{l}
B2:\\
a)x = \dfrac{2}{3}\\
b)x = - 2\\
B3:\\
{x^2} - x + 19 > 0\forall x
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B2:\\
a)4{x^2} - 4x + 1 - 4{x^2} - 8x = - 7\\
\to - 12x = - 8\\
\to x = \dfrac{2}{3}\\
b){x^2} - 1 - {x^2} - 3x = 5\\
\to - 3x = 6\\
\to x = - 2\\
B3:\\
A = {x^2} - x + 19\\
= {x^2} - 2.x.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{{75}}{4}\\
= {\left( {x - \dfrac{1}{2}} \right)^2} + \dfrac{{75}}{4}\\
Do:{\left( {x - \dfrac{1}{2}} \right)^2} \ge 0\forall x\\
\to {\left( {x - \dfrac{1}{2}} \right)^2} + \dfrac{{75}}{4} \ge \dfrac{{75}}{4}\\
\to {x^2} - x + 19 > 0\forall x
\end{array}\)
