Bài 4 trang 76 SGK Đại số và Giải tích 12 Nâng cao, Thực hiện phép tính:...

Bài 4. Thực hiện phép tính:

a) ({81^{ – 0,75}} + {left( {{1 over {125}}} ight)^{{{ – 1} over 3}}} – {left( {{1 over {32}}} ight)^{{{ – 3} over 5}}};) 

b) (0,{001^{{{ – 1} over 3}}} – {left( { – 2} ight)^{ – 2}}{.64^{{2 over 3}}} – {8^{ – 1{1 over 3}}} + {left( {{9^o}} ight)^2};)

c) ({27^{{2 over 3}}} + {left( {{1 over {16}}} ight)^{ – 0,75}} – {25^{0,5}}) 

d) ({left( { – 0,5} ight)^{ – 4}} – {625^{0,25}} – {left( {2{1 over 4}} ight)^{ – 1{1 over 2}}} + 19{left( { – 3} ight)^{ – 3}})

Giải

a) ({81^{ – 0,75}} + {left( {{1 over {125}}} ight)^{{{ – 1} over 3}}} – {left( {{1 over {32}}} ight)^{{{ – 3} over 5}}} = {left( {{3^4}} ight)^{  {{ – 3} over 4}}} + {left( {{{left( {{1 over 5}} ight)}^3}} ight)^{{{ – 1} over 3}}} – {left( {{{left( {{1 over 2}} ight)}^5}} ight)^{{{ – 3} over 5}}})

(, = {left( 3 ight)^{ – 3}} + {left( {{1 over 5}} ight)^{ – 1}} – {left( {{1 over 2}} ight)^{ – 3}} = {1 over {27}} + 5 – 8 = {1 over {27}} – 3 =  – {{80} over {27}})

b) (0,{001^{{{ – 1} over 3}}} – {left( { – 2} ight)^{ – 2}}{.64^{{2 over 3}}} – {8^{ – 1{1 over 3}}} + {left( {{9^o}} ight)^2} = {left( {{{10}^{ – 3}}} ight)^{ – {1 over 3}}} – {2^{ – 2}}.{left( {{2^6}} ight)^{{2 over 3}}} – {left( {{2^3}} ight)^{ – {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}}} ight)^{ – 0,75}} – {25^{0,5}} = {left( {{3^3}} ight)^{{2 over 3}}} + {left( {{2^{ – 4}}} ight)^{ – {3 over 4}}} – {left( {{5^2}} ight)^{{1 over 2}}} = {3^2} + {2^3} – 5 = 12)

d) ({left( { – 0,5} ight)^{ – 4}} – {625^{0,25}} – {left( {2{1 over 4}} ight)^{ – 1{1 over 2}}} + 19{left( { – 3} ight)^{ – 3}} = {left( {{{left( { – 2} ight)}^{ – 1}}} ight)^{ – 4}} – {left( {{5^4}} ight)^{{1 over 4}}} – {left( {{{left( {{3 over 2}} ight)}^2}} ight)^{ – {3 over 2}}} + {{19} over { – 27}})

( = {2^4} – 5 – {left( {{3 over 2}} ight)^{ – 3}} – {{19} over {27}} = 11 – {8 over {27}} – {{19} over {27}} = 10.)