Đáp án:
1) a) \(f\left( x \right) > 0\forall x\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\\
a)f\left( x \right) = 4{x^2} - 7x + 8\\
= 4{x^2} - 2.2x.\dfrac{7}{4} + \dfrac{{49}}{{16}} + \dfrac{{79}}{{16}}\\
= {\left( {2x - \dfrac{7}{4}} \right)^2} + \dfrac{{79}}{{16}}\\
Do:{\left( {2x - \dfrac{7}{4}} \right)^2} + \dfrac{{79}}{{16}} > 0\forall x\\
\to f\left( x \right) > 0\forall x\\
b)f\left( x \right) = - 6{x^2} - 7x - 10\\
= - \left( {6{x^2} + 2.x\sqrt 6 .\dfrac{7}{{2\sqrt 6 }} + \dfrac{{49}}{{24}} + \dfrac{{191}}{{24}}} \right)\\
= - {\left( {x\sqrt 6 + \dfrac{7}{{2\sqrt 6 }}} \right)^2} - \dfrac{{191}}{{24}}\\
Do: - {\left( {x\sqrt 6 + \dfrac{7}{{2\sqrt 6 }}} \right)^2} - \dfrac{{191}}{{24}} < 0\forall x\\
\to f\left( x \right) < 0\forall x\\
c)f\left( x \right) = {x^2} + 9 > 0\forall x\\
\to f\left( x \right) > 0\forall x\\
2)a) - 3{x^2} + 5x - 2 \ge 0\\
\to \left( {1 - x} \right)\left( {3x - 2} \right) \ge 0\\
\to \left[ \begin{array}{l}
x \ge 1\\
x \le \dfrac{2}{3}
\end{array} \right.\\
b){x^2} + x - 6 \ge 0\\
\to \left( {x + 3} \right)\left( {x - 2} \right) \ge 0\\
\to \left[ \begin{array}{l}
x \ge 2\\
x \le - 3
\end{array} \right.
\end{array}\)
