Question

b ) \(\frac{1-3x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2}{2x-4x^2}\)

Answer 1

\(\frac{1-3x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2}{2x-4x^2}\)

\(=\frac{\left(1-3x\right)\left(2x-1\right)+2x\left(3x-2\right)+2-3x}{2x\left(2x-1\right)}\)

\(=\frac{2x-1-6x^2+3x+6x^2-4x+2-3x}{2x\left(2x-1\right)}\)

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

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

Answer 2

\(\frac{1-3x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2}{2x-4x^2}\)
\(=\frac{\left(1-3x\right)\left(2x-1\right)}{2x\left(2x-1\right)}+\frac{2x\left(3x-2\right)}{2x\left(2x-1\right)}-\frac{3x-2}{2x\left(2x-1\right)}\)
\(=\frac{2x-1-6x^2+3x+6x^2-4x-3x+2}{2x\left(2x-1\right)}\)
\(=\frac{-2x+1}{2x\left(2x-1\right)}\)
\(=\frac{-\left(2x-1\right)}{2x\left(2x-1\right)}\)
\(=\frac{-1}{2x}\)