\(B=\dfrac{2x^2-18x}{x^4-81}=\dfrac{2\left(x^2-9\right)}{x^4-81}=\dfrac{2\left(x-3\right)\left(x+3\right)}{\left(x^2-9\right)\left(x^2+9\right)}=\dfrac{2\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x^2+9\right)}=\dfrac{2}{x^2+9}\)
Cho mình sửa,vừa nãy gõ thiếu chữ x:V
\(B=\dfrac{2x^2-18x}{x^4-81}=\dfrac{2\left(x^2-9x\right)}{x^4-81}=\dfrac{2\left(x-3\right)\left(x+3\right)}{\left(x^2-9\right)\left(x^2+9\right)}=\dfrac{2\left(x-3\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)\left(x^2+9\right)}=\dfrac{2}{x^2+9}\)