Bài 4 trang 76 SGK Đại số và Giải tích 12 Nâng cao

Hướng dẫn giải

a) $${81^{ - 0,75}} + {\left( {{1 \over {125}}} \right)^{{{ - 1} \over 3}}} - {\left( {{1 \over {32}}} \right)^{{{ - 3} \over 5}}} = {\left( {{3^4}} \right)^{ {{ - 3} \over 4}}} + {\left( {{{\left( {{1 \over 5}} \right)}^3}} \right)^{{{ - 1} \over 3}}} - {\left( {{{\left( {{1 \over 2}} \right)}^5}} \right)^{{{ - 3} \over 5}}}$$

$$\, = {\left( 3 \right)^{ - 3}} + {\left( {{1 \over 5}} \right)^{ - 1}} - {\left( {{1 \over 2}} \right)^{ - 3}} = {1 \over {27}} + 5 - 8 = {1 \over {27}} - 3 = - {{80} \over {27}}$$

b) $$0,{001^{{{ - 1} \over 3}}} - {\left( { - 2} \right)^{ - 2}}{.64^{{2 \over 3}}} - {8^{ - 1{1 \over 3}}} + {\left( {{9^o}} \right)^2} = {\left( {{{10}^{ - 3}}} \right)^{ - {1 \over 3}}} - {2^{ - 2}}.{\left( {{2^6}} \right)^{{2 \over 3}}} - {\left( {{2^3}} \right)^{ - {4 \over 3}}} + 1$$

$$= 10 - {2^2} - {2^{ - 4}} + 1 = 7 - {1 \over {16}} = {{111} \over {16}}$$

c) $${27^{{2 \over 3}}} + {\left( {{1 \over {16}}} \right)^{ - 0,75}} - {25^{0,5}} = {\left( {{3^3}} \right)^{{2 \over 3}}} + {\left( {{2^{ - 4}}} \right)^{ - {3 \over 4}}} - {\left( {{5^2}} \right)^{{1 \over 2}}} = {3^2} + {2^3} - 5 = 12$$

d) $${\left( { - 0,5} \right)^{ - 4}} - {625^{0,25}} - {\left( {2{1 \over 4}} \right)^{ - 1{1 \over 2}}} + 19{\left( { - 3} \right)^{ - 3}} = {\left( {{{\left( { - 2} \right)}^{ - 1}}} \right)^{ - 4}} - {\left( {{5^4}} \right)^{{1 \over 4}}} - {\left( {{{\left( {{3 \over 2}} \right)}^2}} \right)^{ - {3 \over 2}}} + {{19} \over { - 27}}$$

$$= {2^4} - 5 - {\left( {{3 \over 2}} \right)^{ - 3}} - {{19} \over {27}} = 11 - {8 \over {27}} - {{19} \over {27}} = 10.$$