f(x)= (-8 x² + x – 3). ( 2x+9) Phần 4 x ² -1 10/11/2021 Bởi Clara f(x)= (-8 x² + x – 3). ( 2x+9) Phần 4 x ² -1
Đáp án: \(\begin{array}{l}f\left( x \right) < 0 \Leftrightarrow x \in \left( { – \dfrac{9}{2}; – \dfrac{1}{2}} \right) \cup \left( {\dfrac{1}{2}; + \infty } \right)\\f\left( x \right) > 0 \Leftrightarrow x \in \left( { – \infty ; – \dfrac{9}{2}} \right) \cup \left( { – \dfrac{1}{2};\dfrac{1}{2}} \right)\end{array}\) Giải thích các bước giải: \(\begin{array}{l}DK:x \ne \pm \dfrac{1}{2}\\f\left( x \right) = 0\\ \to \dfrac{{\left( { – 8{x^2} + x – 3} \right)\left( {2x + 9} \right)}}{{\left( {2x – 1} \right)\left( {2x + 1} \right)}} = 0\\ \to 2x + 9 = 0\\ \to x = – \dfrac{9}{2}\\\left( {do: – 8{x^2} + x – 3 < 0\forall x} \right)\end{array}\) BXD: x -∞ -9/2 -1/2 1/2 +∞ f(x) + 0 – // + // – \(\begin{array}{l}KL:f\left( x \right) < 0 \Leftrightarrow x \in \left( { – \dfrac{9}{2}; – \dfrac{1}{2}} \right) \cup \left( {\dfrac{1}{2}; + \infty } \right)\\f\left( x \right) > 0 \Leftrightarrow x \in \left( { – \infty ; – \dfrac{9}{2}} \right) \cup \left( { – \dfrac{1}{2};\dfrac{1}{2}} \right)\end{array}\) Bình luận
Đáp án:
\(\begin{array}{l}
f\left( x \right) < 0 \Leftrightarrow x \in \left( { – \dfrac{9}{2}; – \dfrac{1}{2}} \right) \cup \left( {\dfrac{1}{2}; + \infty } \right)\\
f\left( x \right) > 0 \Leftrightarrow x \in \left( { – \infty ; – \dfrac{9}{2}} \right) \cup \left( { – \dfrac{1}{2};\dfrac{1}{2}} \right)
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne \pm \dfrac{1}{2}\\
f\left( x \right) = 0\\
\to \dfrac{{\left( { – 8{x^2} + x – 3} \right)\left( {2x + 9} \right)}}{{\left( {2x – 1} \right)\left( {2x + 1} \right)}} = 0\\
\to 2x + 9 = 0\\
\to x = – \dfrac{9}{2}\\
\left( {do: – 8{x^2} + x – 3 < 0\forall x} \right)
\end{array}\)
BXD:
x -∞ -9/2 -1/2 1/2 +∞
f(x) + 0 – // + // –
\(\begin{array}{l}
KL:f\left( x \right) < 0 \Leftrightarrow x \in \left( { – \dfrac{9}{2}; – \dfrac{1}{2}} \right) \cup \left( {\dfrac{1}{2}; + \infty } \right)\\
f\left( x \right) > 0 \Leftrightarrow x \in \left( { – \infty ; – \dfrac{9}{2}} \right) \cup \left( { – \dfrac{1}{2};\dfrac{1}{2}} \right)
\end{array}\)
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