Bài 101277

Question

Giải các bất phương trình :

$\begin{array}{l}
1)\,\,\,{\log _{{x^2}}}\left( {3 - 2x} \right) > 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\\
2)\,\,{\log _x}\frac{3}{{8 - 2x}} > - 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)\\
3)\,\,{\log _{{x^2}}}\frac{{2x}}{{\left| {x - 3} \right|}} \le \frac{1}{{2\,}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(3)
\end{array}$

score 0 · Answer 1

$1)$
Điều kiện: $\left\{ \begin{array}{l} x\neq\pm1\\ x<\frac{3}{2}\\x\neq0 \end{array} \right.$
Khi đó BPT đã cho trở thành
$(1) \Leftrightarrow \left\{ \begin{array}{l}
{x^2} > 1\\
3 - 2x > {x^2}
\end{array} \right.\,\,\,\,\,\,\,(a)$        hoặc   $\left\{ \begin{array}{l}
0 < {x^2} < 1\\
0 < 3 - 2x < {x^2}
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(q)$
$\begin{array}{l}
(a) \Leftrightarrow \left\{ \begin{array}{l} x^2>1\\ x^2+2x-3<0 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} \left[ \begin{array}{l} x>1\\ x<-1 \end{array} \right.\\ -3<x<1 \end{array} \right.\Rightarrow - 3 < x < - 1(TM)\\
(q) \Leftrightarrow \left\{ \begin{array}{l} -1<x<1\\ x^2+2x-3>0 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} -1<x<1\\ \left[\begin{array}{l} x>1\\ x<-3 \end{array} \right. \end{array} \right.\Rightarrow \left\{ \begin{array}{l}
- 1 < x < 1,\,x \ne 0\\
x < - 3\,\,\,;\,\,x > 1
\end{array} \right.
\end{array}$
Vậy $(q)$ vô nghiệm
Tóm lại bất phương trình $(1)$ có nghiệm:
$$ - 3 < x < - 1$$
$2)$
Điều kiện : $\left\{ \begin{array}{l} 0<x\neq1\\ 8-2x>0 \end{array} \right.\Rightarrow \left\{ \begin{array}{l} 0<x\neq1\\ x<4 \end{array} \right.$
Khi đó BPT đã cho tương đương
$\left[ \begin{array}{l} \left\{ \begin{array}{l} 4>x>1\\ \frac{3}{8-2x}>\frac{1}{x^2} \end{array} \right.(I)\\ \left\{ \begin{array}{l} 0<x<1\\ \frac{3}{8-2x}<\frac{1}{x^2}  \end{array} \right.(II) \end{array} \right.$
Ta có
$(I)\Leftrightarrow \left\{ \begin{array}{l} 1<x<4\\ 3x^2+2x-8>0 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} 1<x<4\\ (3x-4)(x+2)>0 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} 1<x<4 \\ \left[ \begin{array}{l} x > \frac{4}{3}\\ x < -2 \end{array} \right. \end{array} \right.$
$\Rightarrow \frac{4}{3} < x < 4 $ .
$(II)\Leftrightarrow \left\{ \begin{array}{l} 0<x<1 \\ 3x^2+2x-8 < 0  \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} 0<x<1 \\ (3x-4)(x+2) < 0  \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} 0<x<1 \\ -2 < x < \frac{4}{3} \end{array} \right.$
$\Rightarrow 0<x<1 $
Tóm lại bất phương trình $(2)$ có nghiệm:       $$\left[ \begin{array}{l}
0 < x < 1\\
\frac{4}{3} < x < 4
\end{array} \right.$$
$3)$
Điều kiện : $\left\{ \begin{array}{l} x\notin\left\{ {0;1} \right\}\\ x>0 \end{array} \right.$
Khi đó BPT đã cho tương đương
$ \log_ { x^2 }\frac{4x^2 }{(x-3)^2}\leq 1 $
$\Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x>3\\ \frac{4x^2 }{(x-3)^2}\leq {x^2} \end{array} \right.(I)\\ \left\{ \begin{array}{l} 1<x<3\\ \frac{4x^2 }{(x-3)^2}\leq {x^2} \end{array} \right.(II) \\ \left\{ \begin{array}{l} 0<x<1\\ \frac{4x^2 }{(x-3)^2}\geq {x^2} \end{array} \right.(III)  \end{array} \right. $
$(I)\Leftrightarrow \left\{ \begin{array}{l} 3<x\\ x(x-3)\geq 2x \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} 3<x \\x(x-5)\geq 0 \end{array} \right.\Leftrightarrow x\geq 5.$
$(II)\Leftrightarrow \left\{ \begin{array}{l} 1<x<3 \\ x(3-x)\geq 2x \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} 1<x<3 \\x(x-1)\leq 0 \end{array} \right.\Leftrightarrow\left\{ \begin{array}{l} 1<x<3 \\ 0\leq x\leq 1\end{array} \right.(L)$
$(III)\Leftrightarrow \left\{ \begin{array}{l}  0<x<1 \\ x(3-x)\leq 2x \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l}   0<x<1 \\x(x-1)\geq  0 \end{array} \right.\Leftrightarrow\left\{ \begin{array}{l} 0<x<1\\ x\notin\left[ {0;1} \right] \end{array} \right.(L)$
Tóm lại nghiệm BPT đã cho là $x \ge 5$

Bài 101277

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Bài 101277

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