Toán Giải bất pt x^4 -5x^2+2x+3 bé hơn hoặc bằng 0 26/09/2021 By Cora Giải bất pt x^4 -5x^2+2x+3 bé hơn hoặc bằng 0
Giải thích các bước giải: $\begin{array}{l}{x^4} – 5{x^2} + 2x + 3 \le 0\\ \Leftrightarrow {x^4} – {x^3} – {x^2} + {x^3} – {x^2} – x – 3{x^2} + 3x + 3 \le 0\\ \Leftrightarrow {x^2}\left( {{x^2} – x – 1} \right) + x\left( {{x^2} – x – 1} \right) – 3\left( {{x^2} – x – 1} \right) \le 0\\ \Leftrightarrow \left( {{x^2} – x – 1} \right)\left( {{x^2} + x – 3} \right) \le 0\\ \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}{x^2} – x – 1 \le 0\\{x^2} + x – 3 \ge 0\end{array} \right.\\\left\{ \begin{array}{l}{x^2} – x – 1 \ge 0\\{x^2} + x – 3 \le 0\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}\dfrac{{1 – \sqrt 5 }}{2} \le x \le \dfrac{{1 + \sqrt 5 }}{2}\\\left[ \begin{array}{l}x \ge \dfrac{{ – 1 + \sqrt {13} }}{2}\\x \le \dfrac{{ – 1 – \sqrt {13} }}{2}\end{array} \right.\end{array} \right.\\\left\{ \begin{array}{l}\left[ \begin{array}{l}x \ge \dfrac{{1 + \sqrt 5 }}{2}\\x \le \dfrac{{1 – \sqrt 5 }}{2}\end{array} \right.\\\dfrac{{ – 1 – \sqrt {13} }}{2} \le x \le \dfrac{{ – 1 + \sqrt {13} }}{2}\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\dfrac{{ – 1 + \sqrt {13} }}{2} \le x \le \dfrac{{1 + \sqrt 5 }}{2}\\\dfrac{{ – 1 – \sqrt {13} }}{2} \le x \le \dfrac{{1 – \sqrt 5 }}{2}\end{array} \right.\end{array}$ Vậy bất phương trình có tập nghiệm là: $S = \left[ {\dfrac{{ – 1 + \sqrt {13} }}{2};\dfrac{{1 + \sqrt 5 }}{2}} \right] \cup \left[ {\dfrac{{ – 1 – \sqrt {13} }}{2};\dfrac{{1 – \sqrt 5 }}{2}} \right]$ Trả lời
Giải thích các bước giải:
$\begin{array}{l}
{x^4} – 5{x^2} + 2x + 3 \le 0\\
\Leftrightarrow {x^4} – {x^3} – {x^2} + {x^3} – {x^2} – x – 3{x^2} + 3x + 3 \le 0\\
\Leftrightarrow {x^2}\left( {{x^2} – x – 1} \right) + x\left( {{x^2} – x – 1} \right) – 3\left( {{x^2} – x – 1} \right) \le 0\\
\Leftrightarrow \left( {{x^2} – x – 1} \right)\left( {{x^2} + x – 3} \right) \le 0\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
{x^2} – x – 1 \le 0\\
{x^2} + x – 3 \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
{x^2} – x – 1 \ge 0\\
{x^2} + x – 3 \le 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
\dfrac{{1 – \sqrt 5 }}{2} \le x \le \dfrac{{1 + \sqrt 5 }}{2}\\
\left[ \begin{array}{l}
x \ge \dfrac{{ – 1 + \sqrt {13} }}{2}\\
x \le \dfrac{{ – 1 – \sqrt {13} }}{2}
\end{array} \right.
\end{array} \right.\\
\left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge \dfrac{{1 + \sqrt 5 }}{2}\\
x \le \dfrac{{1 – \sqrt 5 }}{2}
\end{array} \right.\\
\dfrac{{ – 1 – \sqrt {13} }}{2} \le x \le \dfrac{{ – 1 + \sqrt {13} }}{2}
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{{ – 1 + \sqrt {13} }}{2} \le x \le \dfrac{{1 + \sqrt 5 }}{2}\\
\dfrac{{ – 1 – \sqrt {13} }}{2} \le x \le \dfrac{{1 – \sqrt 5 }}{2}
\end{array} \right.
\end{array}$
Vậy bất phương trình có tập nghiệm là: $S = \left[ {\dfrac{{ – 1 + \sqrt {13} }}{2};\dfrac{{1 + \sqrt 5 }}{2}} \right] \cup \left[ {\dfrac{{ – 1 – \sqrt {13} }}{2};\dfrac{{1 – \sqrt 5 }}{2}} \right]$