Phương pháp giải:

- Áp dụng kiến thức các phép toán cộng, trừ, nhân, chia, và thứ tự thực hiện phép toán: nhân chia trước, cộng trừ sau.

- Sử dụng các phép tính về lũy thừa:  ${a^m}.{a^n} = {a^{m + n}};\,\,{a^m}:{a^n} = {a^{m - n}}\,\,\,\left( {m \ge n} \right);\,\,\,{\left( {{a^m}} \right)^n} = {a^{m.n}}.$

Lời giải chi tiết:

$\eqalign{& a)\,\,195 - 3(x - 5) = 12  \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3(x - 5)\, = 195 - 12  \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3(x - 5) = 183  \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x - 5 = 183:3  \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x - 5 = 61  \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\, = 61 + 5  \cr  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\, = 66. \cr}$                  $\eqalign{ & b)\,\,{(x - 4)^8}:{3^2} = {9^3}  \cr  & \,\,\,\,\,\,{(x - 4)^8}\,\,\,\,\,\,\,\,\,\,\, = {({3^2})^3}{.3^2}  \cr & \,\,\,\,\,\,{(x - 4)^8}\,\,\,\,\,\,\,\,\,\,\, = {3^6}{.3^2}  \cr & \,\,\,\,\,\,{(x - 4)^8}\,\,\,\,\,\,\,\,\,\,\, = {3^8}  \cr & \,\,\,\,\,\,\,x - 4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3  \cr & \,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3 + 4  \cr & \,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 7. \cr}$