Giải Bất Phương Trình: 1) (x^2-3x)căn(2x^2-3x-2 Lớn Hơn Bằng 0 2 ...

$\begin{array}{l} 1)\left( {{x^2} - 3x} \right)\sqrt {2{x^2} - 3x - 2} \ge 0\\ \Rightarrow \left\{ \begin{array}{l} {x^2} - 3x \ge 0\\ 2{x^2} - 3x - 2 \ge 0 \end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l} \left[ \begin{array}{l} x \ge 3\\ x \le 0 \end{array} \right.\\ \left[ \begin{array}{l} x \ge 2\\ x \le - \frac{1}{2} \end{array} \right. \end{array} \right. \Rightarrow \left[ \begin{array}{l} x \ge 3\\ x \le - \frac{1}{2} \end{array} \right.\\ 2)\dfrac{{2x}}{{\sqrt {2x + 1} - 1}} > 2x + 2\left( {\text{đkxđ}:x \ge - \dfrac{1}{2};x \ne 0} \right)\\ \Rightarrow \dfrac{{2x\left( {\sqrt {2x + 1} + 1} \right)}}{{\left( {\sqrt {2x + 1} - 1} \right)\left( {\sqrt {2x + 1} + 1} \right)}} > 2x + 2\\ \Rightarrow \dfrac{{2x.\left( {\sqrt {2x + 1} + 1} \right)}}{{2x + 1 - 1}} > 2x + 2\\ \Rightarrow \sqrt {2x + 1} + 1 > 2x + 2\\ \Rightarrow \sqrt {2x + 1} > 2x + 1\\ \Rightarrow 2x + 1 > 4{x^2} + 4x + 1\\ \Rightarrow 4{x^2} + 2x < 0\\ \Rightarrow - \dfrac{1}{2} < x < 0\\ \text{Vậy}\, - \dfrac{1}{2} < x < 0\\ 4)\left( {x + 3} \right).\sqrt {{x^2} - 4} \ge {x^2} - 9\left( {\text{đk}:x \ge 2\text{ hoặc }x \le - 2} \right)\\ \Rightarrow \left( {x + 3} \right).\left( {\sqrt {{x^2} - 4} - x + 3} \right) \ge 0\\ \Rightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x \ge - 3\\ {x^2} - 4 \ge {x^2} - 6x + 9 \end{array} \right.\\ \left\{ \begin{array}{l} x \le - 3\\ {x^2} - 4 \le {x^2} - 6x + 9\\ x \ge 3 \end{array} \right. \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x \ge - 3\\ x \ge \frac{{13}}{6} \end{array} \right. \Rightarrow x \ge \frac{{13}}{6} \end{array}$

$%\begin{array}{l} 3)\frac{{2x}}{{{{\left( {3 - \sqrt {9 + 2x} } \right)}^2}}} \le x + 21\left( {x \ne 0;x \ge - \frac{9}{2}}\right)\\ \Rightarrow \frac{{2x - \left( {x + 21} \right).\left( {9 + 9 + 2x - 6\sqrt {9 + 2x} } \right)}}{{{{\left( {3 - \sqrt {9 + 2x} } \right)}^2}}} \le 0\\ \Rightarrow 2x - \left( {x + 21} \right).\left( {18 + 2x - 6\sqrt {9 + 2x} } \right) \le 0\\ \Rightarrow x - \left( {x + 21} \right).\left( {x + 9 - 3\sqrt {9 + 2x} } \right) \le 0\\\end{array}$