Đáp án:
\(\begin{array}{l}
f\left( x \right) > 0 \Leftrightarrow x \in \left( { - \dfrac{1}{{2\sqrt 2 - 1}}; + \infty } \right)\\
f\left( x \right) < 0 \Leftrightarrow x \in \left( { - \infty ; - \dfrac{1}{{2\sqrt 2 - 1}}} \right)
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
f\left( x \right) = \left( {2\sqrt 2 - 1} \right)x + 1\\
f\left( x \right) = 0\\
\to \left( {2\sqrt 2 - 1} \right)x + 1 = 0\\
\to x = - \dfrac{1}{{2\sqrt 2 - 1}}
\end{array}\)
BXD:
x -∞ \( - \dfrac{1}{{2\sqrt 2 - 1}}\) +∞
f(x) - 0 +
\(\begin{array}{l}
KL:f\left( x \right) > 0 \Leftrightarrow x \in \left( { - \dfrac{1}{{2\sqrt 2 - 1}}; + \infty } \right)\\
f\left( x \right) < 0 \Leftrightarrow x \in \left( { - \infty ; - \dfrac{1}{{2\sqrt 2 - 1}}} \right)
\end{array}\)
