Đáp án: $\dfrac{-16x^{3}+10x^{2}+3x}{4x-2}$
Giải thích các bước giải:
$\left ( 1-\dfrac{2x}{2x} \right )+\dfrac{2x}{2x-1}+\left ( \dfrac{1}{2}x-4x^{2} \right )\\=(1-1)+\dfrac{2x}{2x-1}+\dfrac{1}{2}x-4x^{2}\\=0+\dfrac{2x}{2x-1}+\dfrac{1}{2}x-4x^{2}\\=\dfrac{4x+x(2x-1)-8x^{2}(2x-1)}{2(2x-1)}\\=\dfrac{4x+2x^{2}-x-16x^{3}+8x^{2}}{4x-2}\\=\dfrac{3x+10x^{2}-16x^{3}}{4x-2}\\=\dfrac{-16x^{3}+10x^{2}+3x}{4x-2}$
