Thực hiện phép tính:
a) \({81^{ - 0,75}} + {\left( {{1 \over {125}}} \right)^{{{ - 1} \over 3}}} - {\left( {{1 \over {32}}} \right)^{{{ - 3} \over 5}}};\)
b) \(0,{001^{{{ - 1} \over 3}}} - {\left( { - 2} \right)^{ - 2}}{.64^{{2 \over 3}}} - {8^{ - 1{1 \over 3}}}\) \( + {\left( {{9^0}} \right)^2};\)
c) \({27^{{2 \over 3}}} + {\left( {{1 \over {16}}} \right)^{ - 0,75}} - {25^{0,5}}\)
d) \({\left( { - 0,5} \right)^{ - 4}} - {625^{0,25}} - {\left( {2{1 \over 4}} \right)^{ - 1{1 \over 2}}} \) \(+ 19{\left( { - 3} \right)^{ - 3}}\)
LG a
\({81^{ - 0,75}} + {\left( {{1 \over {125}}} \right)^{{{ - 1} \over 3}}} - {\left( {{1 \over {32}}} \right)^{{{ - 3} \over 5}}};\)
Lời giải chi tiết:
\({81^{ - 0,75}} + {\left( {{1 \over {125}}} \right)^{{{ - 1} \over 3}}} - {\left( {{1 \over {32}}} \right)^{{{ - 3} \over 5}}} \)
\(= {\left( {{3^4}} \right)^{ {{ - 3} \over 4}}} + {\left( {{{\left( {{1 \over 5}} \right)}^3}} \right)^{{{ - 1} \over 3}}} - {\left( {{{\left( {{1 \over 2}} \right)}^5}} \right)^{{{ - 3} \over 5}}}\)
\(\, = {\left( 3 \right)^{ - 3}} + {\left( {{1 \over 5}} \right)^{ - 1}} - {\left( {{1 \over 2}} \right)^{ - 3}}\)
\(= {1 \over {27}} + 5 - 8 = {1 \over {27}} - 3 = - {{80} \over {27}}\)
Cách khác:
LG b
\(0,{001^{{{ - 1} \over 3}}} - {\left( { - 2} \right)^{ - 2}}{.64^{{2 \over 3}}} - {8^{ - 1{1 \over 3}}}\) \( + {\left( {{9^0}} \right)^2};\)
Lời giải chi tiết:
\(0,{001^{{{ - 1} \over 3}}} - {\left( { - 2} \right)^{ - 2}}{.64^{{2 \over 3}}} - {8^{ - 1{1 \over 3}}} + {\left( {{9^0}} \right)^2} \)
\(= {\left( {{{10}^{ - 3}}} \right)^{ - {1 \over 3}}} - {2^{ - 2}}.{\left( {{2^6}} \right)^{{2 \over 3}}} - {\left( {{2^3}} \right)^{ - {4 \over 3}}} + 1\)
\( = 10 - {2^2} - {2^{ - 4}} + 1 = 7 - {1 \over {16}} = {{111} \over {16}}\)
Cách khác:
LG c
\({27^{{2 \over 3}}} + {\left( {{1 \over {16}}} \right)^{ - 0,75}} - {25^{0,5}}\)
Lời giải chi tiết:
\({27^{{2 \over 3}}} + {\left( {{1 \over {16}}} \right)^{ - 0,75}} - {25^{0,5}} \)
\(= {\left( {{3^3}} \right)^{{2 \over 3}}} + {\left( {{2^{ - 4}}} \right)^{ - {3 \over 4}}} - {\left( {{5^2}} \right)^{{1 \over 2}}} \)
\(= {3^2} + {2^3} - 5 = 12\)
Cách khác:
LG d
\({\left( { - 0,5} \right)^{ - 4}} - {625^{0,25}} - {\left( {2{1 \over 4}} \right)^{ - 1{1 \over 2}}} \) \(+ 19{\left( { - 3} \right)^{ - 3}}\)
Lời giải chi tiết:
\({\left( { - 0,5} \right)^{ - 4}} - {625^{0,25}} - {\left( {2{1 \over 4}} \right)^{ - 1{1 \over 2}}} \) \(+ 19{\left( { - 3} \right)^{ - 3}} \)
\(= {\left( {{{\left( { - 2} \right)}^{ - 1}}} \right)^{ - 4}} - {\left( {{5^4}} \right)^{{1 \over 4}}} - {\left( {{{\left( {{3 \over 2}} \right)}^2}} \right)^{ - {3 \over 2}}} \) \(+ {{19} \over { - 27}}\)
\( = {2^4} - 5 - {\left( {{3 \over 2}} \right)^{ - 3}} - {{19} \over {27}} \)
\(= 11 - {8 \over {27}} - {{19} \over {27}} = 10.\)
Cách khác:
