$\dfrac{5x-2}{x^2-4}-\dfrac{3}{x+2}+\dfrac{x}{x-2}$

= $\dfrac{5x-2}{(x+2)(x-2)}-\dfrac{3}{x+2}+\dfrac{x}{x-2}$

= $\dfrac{5x-2}{(x+2)(x-2)}-\dfrac{3(x-2)}{(x+2)(x-2)}+\dfrac{x(x+2)}{(x-2)(x+2)}$

= $\dfrac{(5x-2)-(3x-6)+(x^2+2x)}{(x+2)(x-2)}$

= $\dfrac{5x-2-3x+6+x^2+2x}{(x+2)(x-2)}$

= $\dfrac{x^2+4x+4}{(x+2)(x-2)}$

= $\dfrac{(x+2)^2}{(x+2)(x-2)}$

=  $\dfrac{x+2}{x-2}$

$ Cho  mik  xin  hay  nhất  ạ $

${#Hoàng}$

Đáp án:

Giải thích các bước giải:

 `P=[5x-2]/[x^2 -4] - 3/[x+2] +x/[x-2]`

`= [5x-2]/[(x-2)(x+2)]- 3/[x+2] +x/[x-2]`

`=[5x-2]/[(x-2)(x+2)] -[3(x-2)]/[(x-2)(x+2)] +[x(x+2)]/[(x-2)(x+2)]`

`= [5x-2-3x+6+x^2 +2x]/[(x-2)(x+2)]`

`= [x^2 +(5x-3x+2x)+(-2+6)]/[(x-2)(x+2)]`

`= [x^2 +4x+4]/[(x-2)(x+2)]`

`= [(x+2)^2]/[(x-2)(x+2)]`

`= [x+2]/[x-2]`

(Bài tham khảo)