Phương pháp giải:

Sử dụng các tính chất: \({a^0} = 1\,\,\,\left( {a \in \mathbb{N}*} \right);\,\,\,\,\,\,\,{1^m} = 1\,\,\,\,\left( {m \in \mathbb{N}*} \right).\) 

Lời giải chi tiết:

\(\begin{array}{l}\,\,\,\,\,{x^{99}} = {x^{1999}}\\\,\,\,\,\,{x^{99}} - {x^{1999}} = 0\end{array}\)

Vì \({x^{1999}} = {x^{1900 + 99}} = {x^{1900}}.\,{x^{99}}\)

Do đó: \({x^{99}} - {x^{1900}}.\,{x^{99}} = 0\)  

           \({x^{99}}.\left( {1 - {x^{1900}}} \right) = 0\)

+ TH1: \({x^{99}} = 0\,\, \Rightarrow x = 0\)

+ TH2: \(1 - {x^{1900}} = 0\,\, \Rightarrow {x^{1900}} = 1 \Rightarrow x = 1\)

Vậy \(x \in \left\{ {0;1} \right\}.\)

Chọn B.

Phương pháp giải:

Sử dụng các tính chất: \({a^0} = 1\,\,\,\left( {a \in \mathbb{N}*} \right);\,\,\,\,\,\,\,{1^m} = 1\,\,\,\,\left( {m \in \mathbb{N}*} \right).\) 

Lời giải chi tiết:

\(\begin{array}{l}\,\,{\left( {x - 9} \right)^{25}} = {\left( {x - 9} \right)^{52}}\,\,\,\,\left( {x \ge 9} \right)\\\,\,{\left( {x - 9} \right)^{25}} - {\left( {x - 9} \right)^{52}} = 0\end{array}\)

Vì \({\left( {x - 9} \right)^{52}} = {\left( {x - 9} \right)^{25 + 27}} = {\left( {x - 9} \right)^{25}}.{\left( {x - 9} \right)^{27}}\)

Do đó:

\(\begin{array}{l}{\left( {x - 9} \right)^{25}} - {\left( {x - 9} \right)^{25}}.{\left( {x - 9} \right)^{27}} = 0\\{\left( {x - 9} \right)^{25}}.\left[ {1 - {{\left( {x - 9} \right)}^{27}}} \right] = 0\end{array}\)

+ TH1: \({\left( {x - 9} \right)^{25}} = 0\,\, \Rightarrow x - 9 = 0\,\, \Rightarrow x = 9\)

+ TH2: \(1 - {\left( {x - 9} \right)^{27}} = 0\)

                 \(\begin{array}{l}{\left( {x - 9} \right)^{27}} = 1\\{\left( {x - 9} \right)^{27}} = {1^{27}}\\\,\,\,\,\,\,\,\,x - 9 = 1\\\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\, = 10\end{array}\)

 Vậy \(x \in \left\{ {9\,;\,\,10} \right\}.\)

Chọn A.