Question

E = \(\dfrac{2x-3}{x-3}-\dfrac{x+1}{x+3}+\dfrac{x^2-1}{9-x^2}\)

 

Answer 1

Đk: `x ne -3`.

`E = ((2x-3)(x+3))/((x-3)(x+3)) - ((x-3)(x+1))/((x-3)(x+3)) - (x^2-1)/(x^2-9)`

`= (2x^2 + 3x - 9 - x^2 + 2x + 3 - x^2 + 1)/(x^2-9)`

`= (5x - 5)/(x^2-9)`

`= (5(x-1))/((x-3)(x+3))`

Answer 2

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

\(=\dfrac{2x^2+6x-3x-9-x^2+2x+3-x^2+1}{\left(x-3\right)\left(x+3\right)}\)

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