Question

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

\(\dfrac{2}{x+3}-\dfrac{3}{3-x}+\dfrac{2-5x}{x^2-9}\)

Answer 1

a: Sửa đề: \(\dfrac{1}{2x}+\dfrac{x-2}{3x-3}+\dfrac{x}{x^2-x}\)

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

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

\(=\dfrac{3x-3+2x^2+2x}{6x\left(x-1\right)}=\dfrac{2x^2+5x-3}{6x\left(x-1\right)}\)

b: \(=\dfrac{2}{x+3}+\dfrac{3}{x-3}+\dfrac{2-5x}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{2x-6+3x+9+2-5x}{\left(x-3\right)\left(x+3\right)}=\dfrac{5}{\left(x-3\right)\left(x+3\right)}\)