Question

Đã trả lời bởi chuyên gia

Answer 1

\[\begin{array}{l}
{x^4} - 13{x^2} + 36 \ge 0\\
\Leftrightarrow {x^4} - 9{x^2} - 4{x^2} + 36 \ge 0\\
\Leftrightarrow {x^2}({x^2} - 9) - 4({x^2} - 9) \ge 0\\
\Leftrightarrow ({x^2} - 9)({x^2} - 4) \ge 0\\
\Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{\left\{ \begin{array}{l}
{x^2} - 9 \ge 0\\
{x^2} - 4 \ge 0
\end{array} \right.}\\
{\left\{ \begin{array}{l}
{x^2} - 9 \le 0\\
{x^2} - 4 \le 0
\end{array} \right.}
\end{array}} \right.\\
\Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{\left\{ \begin{array}{l}
{x^2} \ge 9\\
{x^2} \ge 4
\end{array} \right.}\\
{\left\{ \begin{array}{l}
{x^2} \le 9\\
{x^2} \le 4
\end{array} \right.}
\end{array}} \right.\\
\Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{\left\{ {\begin{array}{*{20}{c}}
{\left[ \begin{array}{l}
x \ge 3\\
x \le - 3
\end{array} \right.}\\
{\left[ \begin{array}{l}
x \ge 2\\
x \le - 2
\end{array} \right.}
\end{array}} \right.}\\
{\left\{ \begin{array}{l}
- 3 \le x \le 3\\
- 2 \le x \le 2
\end{array} \right.}
\end{array}} \right.\\
\Leftrightarrow \left[ {\begin{array}{*{20}{c}}
\begin{array}{l}
x \ge 3\\
x \le - 3
\end{array}\\
{ - 2 \le x \le 2}
\end{array}} \right.
\end{array}\]

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Answer 2

\[\begin{array}{l}
{x^4} - 13{x^2} + 36 \ge 0\\
\Leftrightarrow {x^4} - 9{x^2} - 4{x^2} + 36 \ge 0\\
\Leftrightarrow {x^2}({x^2} - 9) - 4({x^2} - 9) \ge 0\\
\Leftrightarrow ({x^2} - 9)({x^2} - 4) \ge 0\\
\Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{\left\{ \begin{array}{l}
{x^2} - 9 \ge 0\\
{x^2} - 4 \ge 0
\end{array} \right.}\\
{\left\{ \begin{array}{l}
{x^2} - 9 \le 0\\
{x^2} - 4 \le 0
\end{array} \right.}
\end{array}} \right.\\
\Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{\left\{ \begin{array}{l}
{x^2} \ge 9\\
{x^2} \ge 4
\end{array} \right.}\\
{\left\{ \begin{array}{l}
{x^2} \le 9\\
{x^2} \le 4
\end{array} \right.}
\end{array}} \right.\\
\Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{\left\{ {\begin{array}{*{20}{c}}
{\left[ \begin{array}{l}
x \ge 3\\
x \le - 3
\end{array} \right.}\\
{\left[ \begin{array}{l}
x \ge 2\\
x \le - 2
\end{array} \right.}
\end{array}} \right.}\\
{\left\{ \begin{array}{l}
- 3 \le x \le 3\\
- 2 \le x \le 2
\end{array} \right.}
\end{array}} \right.\\
\Leftrightarrow \left[ {\begin{array}{*{20}{c}}
\begin{array}{l}
x \ge 3\\
x \le - 3
\end{array}\\
{ - 2 \le x \le 2}
\end{array}} \right.
\end{array}\]

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