\(a,2x+7\ge0\Leftrightarrow2x\ge-7\Rightarrow x\ge\dfrac{-7}{2}\)
\(b,5-2x\le0\Leftrightarrow-2x\le-5\Leftrightarrow x\ge\dfrac{5}{2}\)
\(c,\dfrac{x+2}{x^2+1}\ge0\Leftrightarrow x+2\ge x^2+1\Leftrightarrow x+2-x^2-1\ge0\Leftrightarrow x-x^2+1\ge0\)\(\Leftrightarrow-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{5}{4}\ge0\Leftrightarrow-\left(x-\dfrac{1}{2}\right)^2\ge-\dfrac{5}{4}\Rightarrow\left(x-\dfrac{1}{2}\right)^2\ge\dfrac{5}{4}\)\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}\ge\sqrt{\dfrac{5}{4}}\\x-\dfrac{1}{2}\ge-\sqrt{\dfrac{5}{4}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x\ge\sqrt{\dfrac{5}{4}}+\dfrac{1}{2}\\x\ge-\sqrt{\dfrac{5}{4}}+\dfrac{1}{2}\end{matrix}\right.\)
\(d,\dfrac{x^2+3}{2-x}< 0\Leftrightarrow x^2+3< 2-x\Leftrightarrow x^2+3-2+x\ge0\Leftrightarrow\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{3}{4}\ge0\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2\ge\dfrac{-3}{4}\)( vô lí )
Vậy : BPT trên vô nghiệm
