Đáp án:
c) \(f\left( x \right) > 0 \Leftrightarrow x \in \left( { - \infty ; - 6 - \sqrt {43} } \right) \cup \left( { - 6 + \sqrt {43} ; + \infty } \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
a)f\left( x \right) = 8{x^2} - x + 7 = 8{x^2} - 2.2\sqrt 2 x.\dfrac{7}{{2.2\sqrt 2 }} + \dfrac{{49}}{{32}} + \dfrac{{175}}{{32}}\\
= {\left( {2\sqrt 2 x - \dfrac{7}{{2.2\sqrt 2 }}} \right)^2} + \dfrac{{175}}{{32}} > 0\forall x\\
b)f\left( x \right) = 5{x^2} + 3x + 2 = {\left( {x\sqrt 5 } \right)^2} + 2.x\sqrt 5 .\dfrac{1}{{\sqrt 5 }} + \dfrac{1}{5} + \dfrac{9}{5}\\
= {\left( {x\sqrt 5 + \dfrac{1}{{\sqrt 5 }}} \right)^2} + \dfrac{9}{5} > 0\forall x\\
c)f\left( x \right) = {x^2} + 12x - 7\\
\to {x^2} + 12x - 7 = 0\\
\to \left[ \begin{array}{l}
x = - 6 + \sqrt {43} \\
x = - 6 - \sqrt {43}
\end{array} \right.
\end{array}\)
x -∞ \( - 6 - \sqrt {43} \) \( - 6 + \sqrt {43} \) +∞
f(x) + 0 - 0 +
\(\begin{array}{l}
f\left( x \right) > 0 \Leftrightarrow x \in \left( { - \infty ; - 6 - \sqrt {43} } \right) \cup \left( { - 6 + \sqrt {43} ; + \infty } \right)\\
f\left( x \right) < 0 \Leftrightarrow x \in \left( { - 6 - \sqrt {43} ; - 6 + \sqrt {43} } \right)
\end{array}\)
