Đáp án: $x>-\dfrac14$
Giải thích các bước giải:
Ta có:
$\dfrac{2x^2-x}{x+4x^2}\le\dfrac12$
$\to\dfrac{x(2x-1)}{x(1+4x)}\le\dfrac12$
$\to\dfrac{2x-1}{1+4x}\le\dfrac12$
$\to\dfrac{2x-1}{1+4x}-\dfrac12\le 0$
$\to \dfrac{2(2x-1)-1(1+4x)}{1+4x}\le 0$
$\to \dfrac{(4x-2)-(1+4x)}{1+4x}\le 0$
$\to \dfrac{-3}{1+4x}\le 0$
$\to 1+4x>0$
$\to x>-\dfrac14$