bất phương trình

Question

$\frac{1}{\sqrt{2x^{2}+3x-5}}>\frac{1}{2x-1}$

Gió! · Answer 1 · 11/9/2013 11:12:24 AM

$\Leftrightarrow$

$\left [ \begin{matrix}\left\{ \begin{array}{l} 2x-1<0\\ 2x^{2}+3x-5>0 \end{array} \right. (I)\\ \left\{ \begin{array}{l} 2x-1>0\\ \sqrt{2x^{2}+3x-5}<2x-1(II) \end{array} \right.\end{matrix} \right.$

Giai (I):

$\Leftrightarrow$

$\left\{ \begin{array}{l} x<1/2\\ x\in (-\infty ;-\frac{5}{2})\cup (1;+\infty) \end{array} \right.$

$\Leftrightarrow$

$x\in(-\infty;-\frac{5}{2})$

Giai (II)

$\Leftrightarrow \left\{ \begin{array}{l} \left\{ \begin{array}{l} 2x-1>0\\ 2x^{2}+3x-5>0 \end{array} \right.\\ 2x^{2}+3x-5<(2x-1)^{2} \end{array} \right.$

$\Leftrightarrow$

$\left\{ \begin{array}{l} \left\{ \begin{array}{l} x>1/2\\ x\in (-\infty ;-5/2)\cup (1;+\infty )\end{array} \right.\\ 2x^{2}-7x+6>0\end{array} \right.$

$\Leftrightarrow$