\(\dfrac{3}{x-2}\ge\dfrac{5}{2x-1}.\\ \Leftrightarrow\dfrac{3}{x-2}-\dfrac{5}{2x-1}\ge0.\\ \Leftrightarrow\dfrac{6x-3-5x+10}{\left(x-2\right)\left(2x-1\right)}\ge0.\\ \Leftrightarrow\dfrac{x+7}{\left(x-2\right)\left(2x-1\right)}\ge0.\)
Ta có:
\(x+7=0.\Leftrightarrow x=-7.\\ x-2=0.\Leftrightarrow x=2.\\ 2x-1=0.\Leftrightarrow x=\dfrac{1}{2}.\)
Đặt \(f\left(x\right)=\dfrac{x+7}{\left(x-2\right)\left(2x-1\right)}.\)
Bảng xét dấu:
\(x\) \(-\infty\) \(-7\) \(\dfrac{1}{2}\) \(2\) \(+\infty\)
\(x+7\) - 0 + | + | +
\(x-2\) - | - | - 0 +
\(2x-1\) - | - 0 + | +
\(f\left(x\right)\) - 0 + || - || +
Vậy \(f\left(x\right)\ge0.\Leftrightarrow x\in[-7;\dfrac{1}{2})\cup\left(2;+\infty\right).\)
