Đáp án:
7) \(\left[ \begin{array}{l}
- 2 \le x \le - 1\\
2 \le x \le 3
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
7)DK:\left| {{x^2} - x} \right| \ge 2\\
\to \left[ \begin{array}{l}
{x^2} - x \ge 2\left( {DK:\left[ \begin{array}{l}
x \ge 1\\
x \le 0
\end{array} \right.} \right)\\
{x^2} - x \le - 2\left( l \right)\left( {0 < x < 1} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x \ge 2\\
x \le - 1
\end{array} \right.\\
Đặt:\sqrt {\left| {{x^2} - x} \right| - 2} = t\left( {t \ge 0} \right)\\
\to \left| {{x^2} - x} \right| - 2 = {t^2}\\
\to \left| {{x^2} - x} \right| = {t^2} + 2\\
\to Bpt:{t^2} + 2 + t \ge 8\\
\to \left[ \begin{array}{l}
t \ge 2\\
t \le - 3
\end{array} \right.\\
\to t \ge 2 \to \left| {{x^2} - x} \right| - 2 \ge 4\\
\to \left| {{x^2} - x} \right| \ge - 6\\
\to \left[ \begin{array}{l}
{x^2} - x \ge - 6\left( l \right)\\
{x^2} - x \le 6
\end{array} \right.\\
\to - 2 \le x \le 3\\
KL:\left[ \begin{array}{l}
- 2 \le x \le - 1\\
2 \le x \le 3
\end{array} \right.\\
8)DK:3 \ge x \ge - 1\\
Đặt:\sqrt {x + 1} + \sqrt {3 - x} = t\left( {t \ge 0} \right)\\
\to x + 1 + 2\sqrt {\left( {x + 1} \right)\left( {3 - x} \right)} + 3 - x = {t^2}\\
\to \sqrt {\left( {x + 1} \right)\left( {3 - x} \right)} = \dfrac{{{t^2} - 3}}{2}\\
Bpt \to t + \dfrac{{{t^2} - 3}}{2} \le 2\\
\to {t^2} - 3 + 2t - 4 \le 0\\
\to - 1 - 2\sqrt 2 \le t \le - 1 + 2\sqrt 2 \\
\to 0 \le t \le - 1 + 2\sqrt 2 \\
\to \sqrt {x + 1} + \sqrt {3 - x} \le - 1 + 2\sqrt 2 \\
\to x + 1 + 2\sqrt {\left( {x + 1} \right)\left( {3 - x} \right)} + 3 - x \le 9 - 4\sqrt 2 \\
\to \sqrt {\left( {x + 1} \right)\left( {3 - x} \right)} \le 3 - 2\sqrt 2 \\
\to \sqrt {\left( {x + 1} \right)\left( {3 - x} \right)} \le \sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} \\
\to \left( {x + 1} \right)\left( {3 - x} \right) \le \sqrt 2 - 1\\
\to - {x^2} + 2x + 3 \le \sqrt 2 - 1\\
\to \left[ \begin{array}{l}
x \le 1 - \sqrt {5 - \sqrt 2 } \\
x \ge 1 + \sqrt {5 - \sqrt 2 }
\end{array} \right.\\
KL:\left[ \begin{array}{l}
- 1 \le x \le 1 - \sqrt {5 - \sqrt 2 } \\
3 \ge x \ge 1 + \sqrt {5 - \sqrt 2 }
\end{array} \right.
\end{array}\)
