\(2x^2+2x-5=2\left(x^2+x-\dfrac{5}{2}\right)\)
\(=2\left(x^2+x+\dfrac{1}{4}-\dfrac{11}{4}\right)\)
\(=2\left(x+\dfrac{1}{2}\right)^2-\dfrac{11}{2}>=-\dfrac{11}{2}\)
\(\Leftrightarrow\dfrac{1}{2x^2+2x-5}< =-\dfrac{2}{11}\)
=>A>=2/11
Dấu '=' xảy ra khi x=-1/2