Bài 1: Tính:
a) \({\left( {{3 \over 7}} \right)^{21}}:{\left( {{9 \over {49}}} \right)^6}.\)
b) \({\left( { – {1 \over 3}} \right)^7}{\left( { – {1 \over 3}} \right)^9}:{\left[ {{{\left( { – {1 \over 3}} \right)}^3}} \right]^5} \)\(\;+ {\left( { – 2} \right)^{12}}.{\left( { – 2} \right)^3}:{\left( { – 2} \right)^{15}}.\)
Bài 2: Chứng minh rằng: \({{{{\left( {{5^4} – {5^3}} \right)}^3}} \over {{{125}^4}}} = {{64} \over {125}}.\)
Bài 3: So sánh: \({\left( {{1 \over 2}} \right)^4}\) và \({\left( {{1 \over 4}} \right)^4}\)
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Bài 1: a) \({\left( {{3 \over 7}} \right)^{21}}:{\left( {{9 \over {49}}} \right)^6} = {\left( {{3 \over 7}} \right)^{21}}:{\left[ {{{\left( {{3 \over 7}} \right)}^2}} \right]^6} \)
\(= {\left( {{3 \over 7}} \right)^{21}}:{\left( {{3 \over 7}} \right)^{12}} = {\left( {{3 \over 7}} \right)^9}.\)
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b) \({\left( { – {1 \over 3}} \right)^7}{\left( { – {1 \over 3}} \right)^9}:{\left[ {{{\left( { – {1 \over 3}} \right)}^3}} \right]^5} \)\(\;+ {\left( { – 2} \right)^{12}}.{\left( { – 2} \right)^3}:{\left( { – 2} \right)^{15}}\)
\( = {\left( { – {1 \over 3}} \right)^{16}}:{\left( { – {1 \over 3}} \right)^{15}} + {\left( { – 2} \right)^{15}}:{\left( { – 2} \right)^{15}} \)
\(= \left( { – {1 \over 3}} \right) + 1 = {2 \over 3}.\)
Bài 2: Biến đổi vế trái ta có:
\({{{{\left( {{5^4} – {5^3}} \right)}^3}} \over {{{125}^4}}} = {{64} \over {125}} = {{{{\left[ {{5^3}\left( {5 – 1} \right)} \right]}^3}} \over {{{\left( {{5^3}} \right)}^4}}} = {{{5^9}{{.4}^3}} \over {{5^{12}}}} \)\(\;= {{{4^3}} \over {{5^3}}} = {{64} \over {125}}\)
Bài 3: Ta có:
\({\left( {{1 \over 4}} \right)^4} = {\left[ {{{\left( {{1 \over 2}} \right)}^2}} \right]^4} = {\left( {{1 \over 2}} \right)^8} = {1 \over {{2^8}}}\)
\(({\left( {{1 \over 2}} \right)^4} = {1 \over {{2^4}}}\)). Vì \({1 \over {{2^8}}} < {1 \over {{2^4}}}\) nên \({\left( {{1 \over 2}} \right)^4} > {\left( {{1 \over 4}} \right)^4}\).
