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Thực hiện phép tính :
a) \(4{1 \over 2}:\left( {2,5 – 3{3 \over 4}} \right) + {\left( {{{ – 1} \over 2}} \right)^2}\)
b) \(1{{13} \over {15}}.0,75 – \left( {{8 \over {15}} + 25\% } \right)\)
c) \(0,75 – {{43} \over {80}}:\left( {{{ – 4} \over 5} + 2,5.{3 \over 4}} \right)\)
d) \(25\% – 1{1 \over 2} + 0,5.{{12} \over 5}\)
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e) \(\left( { – 3,2} \right).{{ – 15} \over {64}} + \left( {0,8 – 2{4 \over {15}}} \right):3{1 \over 2}\).
\(\eqalign{ & a)4{1 \over 2}:\left( {2,5 – 3{3 \over 4}} \right) + {\left( {{{ – 1} \over 2}} \right)^2} = {9 \over 2}:\left( {{5 \over 2} – {{15} \over 4}} \right) + {1 \over 4} \cr & = {9 \over 2}:\left( {{{10} \over 4} – {{15} \over 4}} \right) + {1 \over 4} = {9 \over 2}:{{ – 5} \over 4} + {1 \over 4} = {9 \over 2}.{{ – 4} \over 5} + {1 \over 4} \cr & = {{ – 18} \over 5} + {1 \over 4} = {{ – 72} \over {20}} + {5 \over {20}} = {{ – 67} \over {20}} = – 3{7 \over {20}}. \cr & b)1{{13} \over {15}}.0,75 – \left( {{8 \over {15}} + 25\% } \right) = {{28} \over {15}}.{3 \over 4} – \left( {{8 \over {15}} + {1 \over 4}} \right) = {7 \over 5} – \left( {{{32} \over {60}} + {{15} \over {60}}} \right) \cr & = {7 \over 5} – {{47} \over {60}} = {{84} \over {60}} – {{47} \over {60}} = {{37} \over {60}}. \cr & c)0,75 – {{43} \over {80}}:\left( {{{ – 4} \over 5} + 2,5.{3 \over 4}} \right) = {3 \over 4} – {{43} \over {80}}:\left( {{{ – 4} \over 5} + {5 \over 2}.{3 \over 4}} \right) \cr & = {3 \over 4} – {{43} \over {80}}:\left( {{{ – 4} \over 5} + {{15} \over 8}} \right) = {3 \over 4} – {{43} \over {80}}:\left( {{{ – 32} \over {40}} + {{75} \over {40}}} \right) = {3 \over 4} – {{43} \over {80}}:{{43} \over {40}} \cr & = {3 \over 4} – {{43} \over {80}}.{{40} \over {43}} = {3 \over 4} – {1 \over 2} = {3 \over 4} – {2 \over 4} = {1 \over 4}. \cr & d)25\% – 1{1 \over 2} + 0,5{{12} \over 5} = {1 \over 4} – {3 \over 2} + {1 \over 2}.{{12} \over 5} \cr & = {1 \over 4} – {3 \over 2} + {6 \over 5} = {5 \over {20}} – {{30} \over {20}} + {{24} \over {20}} = {{ – 1} \over {20}} \cr & e)( – 3,2).{{ – 15} \over {64}} + \left( {0,8 – 2{4 \over {15}}} \right):3{1 \over 2} = – {{16} \over 5}.{{ – 15} \over {64}} + \left( {{4 \over 5} – {{34} \over {15}}} \right):{7 \over 2} \cr & = {3 \over 4} + \left( {{{12} \over {15}} – {{34} \over {15}}} \right):{7 \over 2} = {3 \over 4} + {{ – 22} \over {15}}.{2 \over 7} = {3 \over 4} + {{ – 44} \over {105}} = {{315} \over {420}} + {{ – 176} \over {420}} = {{139} \over {420}}. \cr} \)