(x ² +3) .(x+7) <0 (x ² + 14).(x ² - 3) >0 04/11/2021 Bởi Allison (x ² +3) .(x+7) <0 (x ² + 14).(x ² - 3) >0
Giải thích các bước giải: $\begin{array}{l}a)\left( {{x^2} + 3} \right)\left( {x + 7} \right) < 0\\ \Leftrightarrow x + 7 < 0\left( {do:{x^2} + 3 \ge 3 > 0,\forall x} \right)\\ \Leftrightarrow x < – 7\\b)\left( {{x^2} + 14} \right)\left( {{x^2} – 3} \right) > 0\left( {do:{x^2} + 14 \ge 14 > 0,\forall x} \right)\\ \Leftrightarrow {x^2} – 3 > 0\\ \Leftrightarrow \left( {x – \sqrt 3 } \right)\left( {x + \sqrt 3 } \right) > 0\\ \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x – \sqrt 3 > 0\\x + \sqrt 3 > 0\end{array} \right.\\\left\{ \begin{array}{l}x – \sqrt 3 < 0\\x + \sqrt 3 < 0\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x > \sqrt 3 \\x > – \sqrt 3 \end{array} \right.\\\left\{ \begin{array}{l}x < \sqrt 3 \\x < – \sqrt 3 \end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x > \sqrt 3 \\x < – \sqrt 3 \end{array} \right.\end{array}$ Bình luận
Giải thích các bước giải:
$\begin{array}{l}
a)\left( {{x^2} + 3} \right)\left( {x + 7} \right) < 0\\
\Leftrightarrow x + 7 < 0\left( {do:{x^2} + 3 \ge 3 > 0,\forall x} \right)\\
\Leftrightarrow x < – 7\\
b)\left( {{x^2} + 14} \right)\left( {{x^2} – 3} \right) > 0\left( {do:{x^2} + 14 \ge 14 > 0,\forall x} \right)\\
\Leftrightarrow {x^2} – 3 > 0\\
\Leftrightarrow \left( {x – \sqrt 3 } \right)\left( {x + \sqrt 3 } \right) > 0\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x – \sqrt 3 > 0\\
x + \sqrt 3 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x – \sqrt 3 < 0\\
x + \sqrt 3 < 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > \sqrt 3 \\
x > – \sqrt 3
\end{array} \right.\\
\left\{ \begin{array}{l}
x < \sqrt 3 \\
x < – \sqrt 3
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x > \sqrt 3 \\
x < – \sqrt 3
\end{array} \right.
\end{array}$