![](https://blog.hocmai.vn/wp-content/uploads/2017/07/hot.gif)
![](https://blog.hocmai.vn/wp-content/uploads/2017/07/hot.gif)
Giải các bất phương trình sau:
1, $ 2x^{2}+\sqrt{x+2}+5 \leq \sqrt{2}(\sqrt{x+2}+x)\sqrt{x^{2}-x+3}+x $
2, $ (x+3)\sqrt{x+2} \geq \frac{x^{3}+4x^{2}+3x-2(x+3)\sqrt[3]{2x+3}}{\sqrt[3]{2x+3}-3} $
3, $ (x^{2}-x-6)\sqrt{x-1}+(x-2)\sqrt{x+1} \geq 3x^{2}-9x+2 $
4, $ \frac{5x-13-\sqrt{57+10x-3x^{2}}}{\sqrt{x+3}-\sqrt{19-3x}}+2\sqrt{x+3} \geq x^{2}+2x+9 $
1, $ 2x^{2}+\sqrt{x+2}+5 \leq \sqrt{2}(\sqrt{x+2}+x)\sqrt{x^{2}-x+3}+x $
2, $ (x+3)\sqrt{x+2} \geq \frac{x^{3}+4x^{2}+3x-2(x+3)\sqrt[3]{2x+3}}{\sqrt[3]{2x+3}-3} $
3, $ (x^{2}-x-6)\sqrt{x-1}+(x-2)\sqrt{x+1} \geq 3x^{2}-9x+2 $
4, $ \frac{5x-13-\sqrt{57+10x-3x^{2}}}{\sqrt{x+3}-\sqrt{19-3x}}+2\sqrt{x+3} \geq x^{2}+2x+9 $