Toán Lớp 10: Giải bất pt: 2/(x³-3x+2) > hoặc = 3/(x-1)
Giúp mk với ạ, mình cảm ơn
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Toán Lớp 10: Giải bất pt: 2/(x³-3x+2) > hoặc = 3/(x-1)
Giúp mk với ạ, mình cảm ơn
Comments ( 1 )
x \ge \dfrac{{ – 3 + \sqrt {105} }}{6}\\
\dfrac{{ – 3 – \sqrt {105} }}{6} \le x < – 2
\end{array} \right.$
Dkxd:x\# 1\\
\dfrac{2}{{{x^3} – 3x + 2}} \ge \dfrac{3}{{x – 1}}\\
\Leftrightarrow \dfrac{2}{{{x^3} – {x^2} + {x^2} – x – 2x + 2}} – \dfrac{3}{{x – 1}} \ge 0\\
\Leftrightarrow \dfrac{2}{{\left( {x – 1} \right)\left( {{x^2} + x – 2} \right)}} – \dfrac{3}{{x – 1}} \ge 0\\
\Leftrightarrow \dfrac{2}{{{{\left( {x – 1} \right)}^2}\left( {x + 2} \right)}} – \dfrac{3}{{x – 1}} \ge 0\\
\Leftrightarrow \dfrac{{2 – 3\left( {x – 1} \right)\left( {x + 2} \right)}}{{{{\left( {x – 1} \right)}^2}\left( {x + 2} \right)}} \ge 0\\
\Leftrightarrow \dfrac{{2 – 3{x^2} – 3x + 6}}{{\left( {x + 2} \right)}} \ge 0\left( {do:{{\left( {x – 1} \right)}^2} > 0} \right)\\
\Leftrightarrow \dfrac{{3{x^2} + 3x – 8}}{{x + 2}} \ge 0\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
3{x^2} + 3x – 8 \ge 0\\
x + 2 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
3{x^2} + 3x – 8 \le 0\\
x + 2 < 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge \dfrac{{ – 3 + \sqrt {105} }}{6}\\
x \le \dfrac{{ – 3 – \sqrt {105} }}{6}
\end{array} \right.\\
x > – 2
\end{array} \right.\\
\left\{ \begin{array}{l}
\dfrac{{ – 3 – \sqrt {105} }}{6} \le x \le \dfrac{{ – 3 + \sqrt {105} }}{6}\\
x < – 2
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x \ge \dfrac{{ – 3 + \sqrt {105} }}{6}\\
\dfrac{{ – 3 – \sqrt {105} }}{6} \le x < – 2
\end{array} \right.\\
Vậy\,\left[ \begin{array}{l}
x \ge \dfrac{{ – 3 + \sqrt {105} }}{6}\\
\dfrac{{ – 3 – \sqrt {105} }}{6} \le x < – 2
\end{array} \right.
\end{array}$