Bài 1: Tính:
a) \({\left( {{1 \over 4}} \right)^{44}}.{\left( {{1 \over 2}} \right)^{12}};\)
b) \({{{3^{17}}{{.81}^{11}}} \over {{{27}^{10}}{{.9}^{15}}}}\)
Bài 2: Chứng minh rằng: \({8^7} – {2^{18}}\) chia hết cho 14.
Bài 3:Tìm \(x \in \mathbb Z\) biết: \({9^{x – 1}} = {1 \over 9}.\)
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Bài 1: a) \({\left( {{1 \over 4}} \right)^{44}}.{\left( {{1 \over 2}} \right)^{12}} = {\left[ {{{\left( {{1 \over 2}} \right)}^2}} \right]^{44}}.{\left( {{1 \over 2}} \right)^{12}}\)\(\; = {\left( {{1 \over 2}} \right)^{88}}.{\left( {{1 \over 2}} \right)^{12}} = {\left( {{1 \over 2}} \right)^{100}}\)
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b) \({{{3^{17}}{{.81}^{11}}} \over {{{27}^{10}}{{.9}^{15}}}} = {{{3^{17}}.{{\left( {{3^4}} \right)}^{11}}} \over {{{\left( {{3^3}} \right)}^{10}}.{{\left( {{3^2}} \right)}^{15}}}} = {{{3^{17}}{{.3}^{44}}} \over {{3^{30}}{{.3}^{30}}}} = {{{3^{61}}} \over {{3^{60}}}} = 3\)
Bài 2: \({8^7} – {2^{18}} = {\left( {{2^3}} \right)^7} – {2^{18}} = {2^{21}} – {2^{18}} \)
\(= {2^{17}}\left( {{2^4} – 2} \right) = {2^{17}}\left( {16 – 2} \right)\)
\( = {2^{17}}.14 \;\vdots\; 14.\)
Bài 3: \({9^{x – 1}} = {1 \over 9} \Rightarrow {9^x}:9 = {1 \over 9} \)
\(\Rightarrow {9^x} = {1 \over 9}.9 \Rightarrow {9^x} = 1 \)
\(\Rightarrow {9^x} = {9^0} \Rightarrow x = 0.\)
