Bài 1:
a) \({\left( {{1 \over 3}} \right)^{ - 1}} - {\left( { - {6 \over 7}} \right)^0} + {\left( { - {1 \over 2}} \right)^2}:2 \)
\(= {1 \over {{1 \over 3}}} - 1 + {1 \over 4}:2\)
\( = 3 - 1 + {1 \over 8} = 2 + {1 \over 8} = {{17} \over 8}.\)
b) \(\left( {1000 - {1^3}} \right)\left( {1000 - {2^3}} \right)...\)\(\;\left( {1000 - {{50}^3}} \right)\)
\( = \left( {1000 - {1^3}} \right)\left( {1000 - {2^3}} \right)...\)\(\;\left( {1000 - {9^3}} \right)\left( {1000 - {{10}^3}} \right)...\)\(\;\left( {1000 - {{50}^3}} \right)\)
Trong tích này có thừa số \(\left( {1000 - {{10}^3}} \right) = 0.\)
Do đó tích trên bằng 0.
Bài 2:
a) \({3^{200}} = {\left( {{3^2}} \right)^{100}} = {9^{100}}\)
\({2^{300}} = {\left( {{2^3}} \right)^{100}} = {8^{100}} < {9^{100}}\). Do đó \({3^{200}}\) > \({2^{300}}.\)
b) \({9^{12}} = {\left( {{3^2}} \right)^{12}} = {3^{24}} = {\left( {{3^3}} \right)^8} = {27^8}\)\(\, > {26^8}\)
Do đó \(({9^{12}}\)) > \({26^8}\).
Bài 3:
\(\left( {\left| x \right| - {1 \over 8}} \right).{\left( { - {1 \over 8}} \right)^5} = {\left( { - {1 \over 8}} \right)^5} \)
\(\Rightarrow \left| x \right| - {1 \over 8} = {\left( { - {1 \over 8}} \right)^7}:{\left( { - {1 \over 8}} \right)^5}\)
\( \Rightarrow \left| x \right| - {1 \over 8} = {\left( { - {1 \over 8}} \right)^2}\)
\(\Rightarrow \left| x \right| - {1 \over 8} = {1 \over {64}} \)
\(\Rightarrow \left| x \right| = {1 \over {64}} + {1 \over 8}\)
\( \Rightarrow \left| x \right| = {9 \over {64}} \)
\(\Rightarrow x = {9 \over {64}}\) hoặc \(x = {{ - 9} \over {64}}.\)
