Bài 101269

Question

Giải các bất phương trình :
$\begin{array}{l}
1){\log _{\frac{1}{2}}}\left( {1 + x - \sqrt {{x^2} - 4} } \right) \le 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3)\,\,{\log _{\frac{1}{5}}}\left( { - x + 1} \right) \le 0\\
2)\,{\log _3}\left( {\sqrt {{x^2} - 9} - x + \frac{1}{3}} \right) \le - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4)\,{\log _2}\left( {2 - x - \sqrt {{x^2} - 1} } \right) \ge 1
\end{array}$

score 0 · Answer 1

$\begin{array}{l}
1)\,\,(1) \Leftrightarrow \left\{ \begin{array}{l}
\left| x \right| \ge 2\\
1 + x - \sqrt {{x^2} - 4} \ge 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left| x \right| \ge 2\\
\sqrt {{x^2} - 4} \le x
\end{array} \right.\\
\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \left\{ \begin{array}{l}
x \ge 2\\
{x^2} - 4 \le {x^2}
\end{array} \right.\\
\Leftrightarrow x\geq 2
\end{array} $

$2)$
$(2)\Leftrightarrow \left\{ \begin{array}{l} |x|\geqslant 3 \\ \sqrt{x^2-9}-x+ \frac{1}{3}\leq \frac{1}{3} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} |x|\geq 3\\ \sqrt{ x^2-9 }\leq x\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x\geq 3\\ x^2-9 \leq x^2\end{array} \right.\\\Leftrightarrow x\geq 3$

$3)$
$3)\Leftrightarrow \left\{ \begin{array}{l} 1-x>0\\ 1-x\geq 1\end{array} \right.\Leftrightarrow x\leq 0$

$4)$
$4) \Leftrightarrow \left\{ \begin{array}{l} x^2-1\geq 0\\ 2-x-\sqrt{x^2-1} \geq 2\end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} |x|\geq 1\\ x+\sqrt{x^2-1} \leq 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} |x|\geq 1\\ x \leq -\sqrt{x^2-1} \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x\leq -1\\ x \leq -\sqrt{x^2-1}(*) \end{array} \right. (**) $
Ta thấy $\forall x<0 (*)\Leftrightarrow -x \geq \sqrt{x^2-1}\Leftrightarrow x^2\geq x^2-1 $ luôn đúng $\forall x<0$ nên
$(**) \Leftrightarrow x \le - 1$

Vậy nghiệm của bpt $x\leq -1 $

Bài 101269

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Bài 101269

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