ĐKXĐ: x<>1/2
Ta có; \(\frac{2x^2+1}{1-2x}\ge1-x\)
=>\(-\frac{2x^2+1}{2x-1}\ge-\left(x-1\right)\)
=>\(\frac{2x^2+1}{2x-1}\le\left(x-1\right)\)
=>\(\frac{2x^2+1-\left(x-1\right)\left(2x-1\right)}{2x-1}\le0\)
=>\(\frac{2x^2+1-\left(2x^2-3x+1\right)}{2x-1}\le0\)
=>\(\frac{3x}{2x-1}\le0\)
=>\(\frac{x}{2x-1}\le0\)
=>\(0\le x<\frac12\)


