Đáp án:
\(\left[ \begin{array}{l}
\dfrac{7}{{\sqrt 2 }} > x \ge - \dfrac{{51}}{{14}}\\
x < - \dfrac{7}{{\sqrt 2 }}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:\left[ \begin{array}{l}
\left\{ \begin{array}{l}
14x + 51 \ge 0\\
49 - 2{x^2} > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
14x + 51 \le 0\\
49 - 2{x^2} < 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge - \dfrac{{51}}{{14}}\\
\dfrac{{49}}{2} > {x^2}
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le - \dfrac{{51}}{{14}}\\
\dfrac{{49}}{2} < {x^2}
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge - \dfrac{{51}}{{14}}\\
\dfrac{7}{{\sqrt 2 }} > x > - \dfrac{7}{{\sqrt 2 }}
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le - \dfrac{{51}}{{14}}\\
\left[ \begin{array}{l}
\dfrac{7}{{\sqrt 2 }} < x\\
x < - \dfrac{7}{{\sqrt 2 }}
\end{array} \right.
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
\dfrac{7}{{\sqrt 2 }} > x \ge - \dfrac{{51}}{{14}}\\
x < - \dfrac{7}{{\sqrt 2 }}
\end{array} \right.
\end{array}\)
