tìm x (3^x-81)(x^2+1)=0

`(3^x-81)(x^2+1)=0`

`=>`$\left[\begin{matrix} 3^x-81=0\\ x^2+1=0\end{matrix}\right.$

`=>`$\left[\begin{matrix} 3^x=81\\ x^2=-1 (vô  lý)\end{matrix}\right.$

`=>3^x=81`

`=>3^x=3^4`

`=>x=4`

Vậy `x=4`