\(\dfrac{3}{x-3}-\dfrac{6x}{x^2-9}=\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{6x}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x+9}{\left(x-3\right)\left(x+3\right)}-\dfrac{6x}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x+9-6x}{\left(x-3\right)\left(x+3\right)}=\dfrac{-3x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{-3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{-3}{x+3}\)
Thực hiện phép tính :
3 6x
___ - ________
x- 3 x^2 - 9
= \(\dfrac{3.\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\) - \(\dfrac{6x}{\left(x+3\right)\left(x-3\right)}\)
= \(\dfrac{3x+9-6x}{\left(x-3\right)\left(x+3\right)}=\dfrac{-3x+9}{\left(x+3\right)\left(x-3\right)}=\dfrac{-3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{-3}{x+3}\)
