Đáp án: \(x \in \left( { – \infty ; – \frac{1}{3}} \right) \cup \left( {0;\frac{1}{2}} \right) \cup \left( {8; + \infty } \right)\) Giải thích các bước giải: \(\begin{array}{l}DK:x \ne \left\{ { – \frac{1}{3};\frac{1}{2}} \right\}\\\frac{{x + 2}}{{3x + 1}} < \frac{{x – 2}}{{2x – 1}}\\ \to \frac{{\left( {x + 2} \right)\left( {2x – 1} \right) – \left( {x – 2} \right)\left( {3x + 1} \right)}}{{\left( {3x + 1} \right)\left( {2x – 1} \right)}} < 0\\ \to \frac{{2{x^2} + 3x – 2 – 3{x^2} + 5x + 2}}{{\left( {3x + 1} \right)\left( {2x – 1} \right)}} < 0\\ \to \frac{{ – {x^2} + 8x}}{{\left( {3x + 1} \right)\left( {2x – 1} \right)}} < 0\\Xét: – {x^2} + 8x = 0\\ \to \left[ \begin{array}{l}x = 0\\x = 8\end{array} \right.\end{array}\) BXD: x -∞ -1/3 0 1/2 8 +∞ f(x) – // + 0 – // + 0 – \(KL:x \in \left( { – \infty ; – \frac{1}{3}} \right) \cup \left( {0;\frac{1}{2}} \right) \cup \left( {8; + \infty } \right)\) Bình luận
Đáp án:
\(x \in \left( { – \infty ; – \frac{1}{3}} \right) \cup \left( {0;\frac{1}{2}} \right) \cup \left( {8; + \infty } \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne \left\{ { – \frac{1}{3};\frac{1}{2}} \right\}\\
\frac{{x + 2}}{{3x + 1}} < \frac{{x – 2}}{{2x – 1}}\\
\to \frac{{\left( {x + 2} \right)\left( {2x – 1} \right) – \left( {x – 2} \right)\left( {3x + 1} \right)}}{{\left( {3x + 1} \right)\left( {2x – 1} \right)}} < 0\\
\to \frac{{2{x^2} + 3x – 2 – 3{x^2} + 5x + 2}}{{\left( {3x + 1} \right)\left( {2x – 1} \right)}} < 0\\
\to \frac{{ – {x^2} + 8x}}{{\left( {3x + 1} \right)\left( {2x – 1} \right)}} < 0\\
Xét: – {x^2} + 8x = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = 8
\end{array} \right.
\end{array}\)
BXD:
x -∞ -1/3 0 1/2 8 +∞
f(x) – // + 0 – // + 0 –
\(KL:x \in \left( { – \infty ; – \frac{1}{3}} \right) \cup \left( {0;\frac{1}{2}} \right) \cup \left( {8; + \infty } \right)\)