1. \(\sqrt{2x^2+5x-6}>2-x\)x
2.\(\sqrt{x^2+2}\le x-1\)
3.\(\sqrt{x^2-2x-15}>2x+5\)
4.\(\left(16-x^2\right)\sqrt{x-3}\le0\)
5.\(\sqrt{x^2+2017}\le\sqrt{2018}x\)
6.\(\hept{\begin{cases}\frac{x+3}{2x-3}-\frac{x}{2x-1}\le0\\\sqrt{x^2+3}+3x< 1\end{cases}}\)
Câu 6:
\(\hept{\begin{cases}\frac{x+3}{2x-3}-\frac{x}{2x-1}\le0\\\sqrt{x^2+3}+3< 1\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{2x^2-x+6x-3-2x^2+3x}{\left(2x-3\right)\left(2x-1\right)}\le0\\x^2+3< \left(1-3x\right)^2\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}8x-3\le0\\x^2+3< 1-6x+9x^2\end{cases}\Leftrightarrow\hept{\begin{cases}8x-3\le0\\8x^2-6x-2< 0\end{cases}\Leftrightarrow}\hept{\begin{cases}x< \frac{3}{8}\\\frac{-1}{4}x< x< \frac{1}{4}\end{cases}\Rightarrow}S\left(\frac{-1}{4};\frac{3}{8}\right)}\)