Bài 101355

Question

Giải các bất phương trình :

$\begin{array}{l}
1)\,\,\,\log _2^2\left( {2 + x - {x^2}} \right) + 3{\log _{\frac{1}{2}}}\left( {2 + x - {x^2}} \right) + 2 \le 0\,\,\,\,\,\,\,\,\,\,\,(1)\\
2)\,\,{\log _{x + 1}}{\left( {{x^2} + x - 6} \right)^2} \ge 4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)\\
3)\,\,\,{\log _{9{x^2}}}\left( {6 + 2x - {x^2}} \right) \le \frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(3)\\
4)\,\,\,{9^{\sqrt {{x^2} - 3} }} + 3 <28. {3^{\sqrt {{x^2} - 3} - 1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(4)
\end{array}$

score 0 · Answer 1

$1)$
Điều kiện: $2+x-x^2>0\Rightarrow -1<x<2.$
Chọn cơ số chung là $2$ ta có:
$(1) \Leftrightarrow \,\,\,\,\,\log _2^2\left( {2 + x - {x^2}} \right) - 3{\log _2}\left( {2 + x - {x^2}} \right) + 2 \le 0\,$
Đặt $t = \log _2^{}\left( {2 + x - {x^2}} \right)$ thì ta có
$t^2-3t+2\leq 0$
$\Leftrightarrow (t-1)(t-2)\leq 0$
$\Leftrightarrow 1\leq t\leq 2$
$\Leftrightarrow 1\leq \log_2(2+x-x^2)\leq 2 $
$\Leftrightarrow 2\leq 2+x-x^2\leq 4$ (do hàm $f(x)=\log_2x$ tăng trên TXĐ)
$\Leftrightarrow \left\{ \begin{array}{l} x^2-x\leq 0\\ x^2-x+2\geq 0 \end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} x(x-1)\leq 0\\ \left ( x-\frac{1}{2} \right )^2+\frac{7}{4}\geq 0 (Đ) \end{array} \right.$
$\Rightarrow 0\leq x\leq 1$ (thỏa tập xác định)
Vậy BPT đã cho có nghiệm : $0 \le x \le 1$

$2)$
Điều kiện: $\left\{ \begin{array}{l} 0<x+1\neq 1\\ x^2+x-6\neq 0 \end{array} \right.\Rightarrow \left\{ \begin{array}{l} -1<x\\ x\notin \left\{ {0;2} \right\} \end{array} \right.$
Khi đó ta có
$(2) \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 1 > 1\\
{\left( {{x^2} + x - 6} \right)^2} \geq {\left( {x + 1} \right)^4}
\end{array} \right.\\
\left\{ \begin{array}{l}
0 < x + 1 < 1\\
0 < {\left( {{x^2} + x - 6} \right)^2} \le {\left( {x + 1} \right)^4}
\end{array} \right.
\end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 2\\
{\left( {{x^2} + x - 6} \right)} \geq {\left( {x + 1} \right)^2}
\end{array} \right.(I)\\
\left\{ \begin{array}{l}
0 < x + 1 < 1\\
{\left( {{x^2} + x - 6} \right)} \geq -{\left( {x + 1} \right)^2}
\end{array} \right.(II)
\end{array} \right. \vee   \left\{ \begin{array}{l} 0<x<2\\
{\left( {{x^2} + x - 6} \right)} \le -{\left( {x + 1} \right)^2}
\end{array} \right. (III)$
$(I)\Leftrightarrow \left\{ \begin{array}{l} x>0\\ x+8\leq 0\end{array} \right. (L)$
$(II)\Leftrightarrow \left\{ \begin{array}{l} -1<x<0\\ 2x^2+3x-5\geq 0\end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} -1<x<0\\ (2x+5)(x-1)\geq 0\end{array} \right.(L)$
$(III)\Leftrightarrow \left\{ \begin{array}{l} 0<x<2\\ 2x^2+3x-5\leq 0  \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 0<x<2\\ (2x+5)(x-1)\leq 0\end{array} \right. \Rightarrow 0<x\le 1(TM)$
Vậy hệ đã cho có nghiệm : $0 < x \le 1$

$3)$
Điều kiện: $\left\{ \begin{array}{l} 0<9x^2\neq 1\\ 6+2x-x^2>0 \end{array} \right.\Rightarrow \left\{ \begin{array}{l} x\notin\left\{ {0;\pm\frac{1}{3}} \right\}\\ 1-\sqrt7<x<1+\sqrt7 \end{array} \right.$
Khi đó ta có
$(3)\Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} 9x^2\geq 1\\x>0\\ 6-2x+x^2\geq 3x \end{array} \right.\vee \left\{ \begin{array}{l} 9x^2> 1\\x\leq 0\\ 6-2x+x^2\leq -3x \end{array} \right.  \\ \left\{ \begin{array}{l} 9x^2\leq 1\\x>0\\ 6-2x+x^2\leq 3x \end{array} \right. \vee \left\{ \begin{array}{l} 9x^2\leq 1\\x\leq 0\\ 6-2x+x^2\geq -3x \end{array} \right. \end{array} \right. $
$ \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x\geq \frac{1}{3}\\ x\geq 5\vee x\leq 1\end{array} \right.\vee \left\{ \begin{array}{l} x\leq -\frac{1}{3} \\ x^2+x+6\leq 0 \end{array} \right. (L) \\  \left\{ \begin{array}{l}  0<x\leq \frac{1}{3} \\ 1\leq x\leq 5\end{array} \right. \vee  \left\{ \begin{array}{l} -\frac{1}{3}\leq x\leq 0\\ x^2+x+6\leq 0  \end{array} \right.  \end{array} \right.  $
$\Leftrightarrow \left[ \begin{array}{l}
1 - \sqrt 7 < x \le - 1\\
- \frac{1}{3} < x < 0\\
0 < x < \frac{1}{3}\\
2 \le x < 1 + \sqrt 7
\end{array} \right.$

$4)$
Điều kiện: $x^2\geq 3$
Đặt $t = {3^{\sqrt {{x^2} - 3} }},\,\,\,{x^2} \ge 3,\,\,t \ge 1$
$(4) \Leftrightarrow {t^2} + 3 < \frac{{28}}{3}t\,\,\,$
$\Leftrightarrow \frac{1}{3} < t < 9$
$\Leftrightarrow -1< \sqrt {{x^2} - 3} <2\Leftrightarrow \left[ \begin{array}{l}
- \sqrt 7 < x \le - \sqrt 3 \\
\sqrt 3 \le x < \sqrt 7
\end{array} \right. $
ĐS :   $\left[ \begin{array}{l}
- \sqrt 7 < x \le - \sqrt 3 \\
\sqrt 3 \le x < \sqrt 7
\end{array} \right.$

Bài 101355

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Bài 101355

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