Bài 1:
a) \({{{4^{20}} - {2^{20}} + {6^{20}}} \over {{6^{20}} - {3^{20}} + {9^{20}}}} = {{{{\left( {{2^2}} \right)}^{20}} - {2^{20}} + {{\left( {2.3} \right)}^{20}}} \over {{{\left( {2.3} \right)}^{20}} - {3^{20}} + {{\left( {{3^2}} \right)}^{20}}}}\)
\(= {{{2^{40}} - {2^{20}} + {2^{20}}{{.3}^{20}}} \over {{2^{20}}{{.3}^{20}} - {3^{20}} + {3^{40}}}}\)\(\; = {{{2^{20}}\left( {{2^{20}} - 1 + {3^{20}}} \right)} \over {{3^{20}}\left( {{2^{20}} - 1 + {3^{20}}} \right)}} = {{{2^{20}}} \over {{3^{20}}}}.\)
b) \({\left( { - 1} \right)^{2n}}{\left( { - 1} \right)^n}{\left( { - 1} \right)^{n + 1}} = {\left( { - 1} \right)^{4n + 1}} \)\(\;= - 1\) (vì \(n \in\mathbb Z\) và \(4n + 1\) là số lẻ).
Bài 2 :
\(2\left| {x - 1} \right| + {\left( { - {1 \over 2}} \right)^5} = {\left( { - {1 \over 4}} \right)^3}\)
\(\Rightarrow 2\left| {x - 1} \right| - {1 \over {{2^5}}} = - {1 \over {{4^3}}}\)
\(\Rightarrow 2\left| {x - 1} \right| = - {1 \over {64}} + {1 \over {32}} \)
\(\Rightarrow 2\left| {x - 1} \right| = {1 \over {64}}.\)
\( \Rightarrow \left| {x - 1} \right| = {1 \over {128}} \)
\(\Rightarrow x - 1 = {1 \over {128}}\) hoặc \(x - 1 = - {1 \over {128}}\)
\( \Rightarrow x = {1 \over {128}} + 1\) hoặc \(x = - {1 \over {128}} + 1\)
\( \Rightarrow x = {{129} \over {128}}\) hoặc \({{127} \over {128}}.\)
