Nghiệm của bất phương trình \(\left(\dfrac{7}{9}\right)^{2x^2-3x}\ge\dfrac{9}{7}\) là
\(x\le\dfrac{1}{2};x\ge1\).\(\dfrac{1}{2}\le x\le1\).\(\dfrac{-3-\sqrt{17}}{4}\le x\le\dfrac{-3+\sqrt{17}}{4}\).\(x\le\dfrac{-3-\sqrt{17}}{4}\) hoặc \(x\ge\dfrac{-3+\sqrt{17}}{4}\).Hướng dẫn giải:\(\left(\dfrac{7}{9}\right)^{2x^2-3x}\ge\dfrac{9}{7}\Leftrightarrow\text{}\left(\dfrac{7}{9}\right)^{2x^2-3x}\ge\left(\dfrac{7}{9}\right)^{-1}\)
\(\Leftrightarrow2x^2-3x\le-1\) (vì \(0< \dfrac{7}{9}< 1\)) \(\Leftrightarrow2x^2-3x+1\le0\) \(\Leftrightarrow\dfrac{1}{2}\le x\le1\).