Question

Tính \(\frac{1-2x}{2x}+\frac{2x}{2x-1}+\frac{1}{2x-4x^2}\)

Answer 1

\(\frac{1-2x}{2x}+\frac{2x}{2x-1}+\frac{1}{2x-4x^2}\)

\(=\frac{1}{2x}-1+1+\frac{1}{2x-1}+\frac{1}{2x\left(1-2x\right)}=\frac{1-2x}{2x\left(1-2x\right)}-\frac{2x}{2x\left(1-2x\right)}+\frac{1}{2x\left(1-2x\right)}\)

\(=\frac{1-2x-2x+1}{2x\left(1-2x\right)}=\frac{2}{2x\left(1-2x\right)}=\frac{1}{x\left(1-2x\right)}\)

Answer 2

Ta có: \(\frac{1-2x}{2x}+\frac{2x}{2x-1}+\frac{1}{2x-4x^2}\)

= \(\frac{1-2x}{2x}+\frac{2x}{2x-1}-\frac{1}{2x\left(2x-1\right)}\)

= \(\frac{\left(1-2x\right)\left(2x-1\right)}{2x\left(2x-1\right)}+\frac{2x.2x}{2x\left(2x-1\right)}-\frac{1}{2x\left(2x-1\right)}\)

= \(\frac{-\left(4x^2-4x+1\right)}{2x\left(2x-1\right)}+\frac{4x^2}{2x\left(2x-1\right)}-\frac{1}{2x\left(2x-1\right)}\)

= \(\frac{-4x^2+4x-1+4x^2-1}{2x\left(2x-1\right)}\)

= \(\frac{4x-2}{2x\left(2x-1\right)}\)

= \(\frac{2\left(2x-1\right)}{2x\left(2x-1\right)}=\frac{1}{x}\)