Bất phương trình \(\left(0,4\right)^{x\left(x+1\right)}>\left(2,5\right)^{3-2x^2}\) có nghiệm là
\(x< \dfrac{-1-\sqrt{13}}{2}\) hoặc \(x>\dfrac{1+\sqrt{13}}{2}\).\(x< \dfrac{1-\sqrt{13}}{2}\) hoặc \(x>\dfrac{1+\sqrt{13}}{2}\).\(\dfrac{1-\sqrt{13}}{2}< x< \dfrac{1+\sqrt{13}}{2}\).\(\dfrac{-1-\sqrt{13}}{2}< x< \dfrac{-1+\sqrt{13}}{2}\).Hướng dẫn giải:Ta có: \(0,4=\dfrac{2}{5}\) , \(2,5=\dfrac{5}{2}\). Bất phương trình đã cho tương đương với \(\left(\dfrac{2}{5}\right)^{x\left(x+1\right)}>\left(\dfrac{5}{2}\right)^{3-2x^2}\)
\(\Leftrightarrow\left(\dfrac{5}{2}\right)^{-x\left(x+1\right)}>\left(\dfrac{5}{2}\right)^{3-2x}\) \(\Leftrightarrow-x\left(x+1\right)>3-2x^2\) \(\Leftrightarrow x^2-x-3>0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x< \dfrac{1-\sqrt{13}}{2}\\x>\dfrac{1+\sqrt{13}}{2}\end{array}\right.\).
