a) \({3 \over 4} + {{ – 5} \over {13}}.{{13} \over {10}}\)
b) \(\left( {{{ – 3} \over 7} + {1 \over 4}} \right):{{15} \over {28}}\)
c) \(2{1 \over 2}:\left( {{5 \over 2} – 3{3 \over 4}} \right) + {\left( {{{ – 1} \over 2}} \right)^2}\)
d) \(\left( {3 – 2{3 \over 5}} \right).4{1 \over 6} – 1{3 \over 5}:1{1 \over {15}}\)
e) \({5 \over 6} + 6{5 \over 6}\left( {11{5 \over {20}} – 9{1 \over 4}} \right):8{1 \over 3}\)
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f) \({{11} \over 9}.{{15} \over 4} – {{11} \over 4}.{7 \over 9} – {{11} \over 9}.{5 \over 4}\)
g) \(\left( { – 3,6 + 5{1 \over 3}:0,4} \right).{{10} \over {73}} – {1 \over 3}:\left( {4,8 – 5{1 \over 2}} \right)\)
\(\eqalign{ & a){3 \over 4} + {{ – 5} \over {13}}.{{13} \over {10}} = {3 \over 4} + {{ – 1} \over 2} = {3 \over 4} + {{ – 2} \over 4} = {1 \over 4} \cr & b)\left( {{{ – 3} \over 7} + {1 \over 4}} \right):{{15} \over {28}} = \left( {{{ – 12} \over {28}} + {7 \over {28}}} \right):{{15} \over {28}} = {{ – 5} \over {28}}.{{28} \over {15}} = {{ – 1} \over 3} \cr & c)2{1 \over 2}:\left( {{5 \over 2} – 3{3 \over 4}} \right) + {\left( {{{ – 1} \over 2}} \right)^2} = {5 \over 2}:\left( {{5 \over 2} – {{15} \over 4}} \right) + {1 \over 4} \cr & = {5 \over 2}:\left( {{{10} \over 4} – {{15} \over 4}} \right) + {1 \over 4} = {5 \over 2}:{{ – 5} \over 4} + {1 \over 4} = {5 \over 2}.{{ – 4} \over 5} + {1 \over 4} = – 2 + {1 \over 4} \cr & = {{ – 8} \over 4} + {1 \over 4} = {{ – 7} \over 4} = – 1{3 \over 4} \cr & d)\left( {3 – 2{3 \over 5}} \right).4{1 \over 6} – 1{3 \over 5}:1{1 \over {15}} = \left( {2{5 \over 5} – 2{3 \over 5}} \right).{{25} \over 6} – {8 \over 5}:{{16} \over {15}} \cr & = {2 \over 5}.{{25} \over 6} – {8 \over 5}.{{15} \over {16}} = {5 \over 3} – {3 \over 2} = {{10} \over 6} – {9 \over 6} = {1 \over 6} \cr & e){5 \over 6} + 6{5 \over 6}\left( {11{5 \over {20}} – 9{1 \over 4}} \right):8{1 \over 3} = {5 \over 6} + {{41} \over 6}\left( {11{5 \over {20}} – 9{5 \over {20}}} \right):{{25} \over 3} \cr & = {5 \over 6} + {{41} \over 6}.2:{{25} \over 3} = {5 \over 6} + {{41} \over 3}.{3 \over {25}} \cr & = {5 \over 6} + {{41} \over 3}.{3 \over {25}} = {5 \over 6} + {{41} \over {25}} = {{125} \over {150}} + {{246} \over {150}} = {{371} \over {150}} = 2{{71} \over {150}} \cr & f){{11} \over 9}.{{15} \over 4} – {{11} \over 4}.{7 \over 9} – {{11} \over 9}.{5 \over 4} = {{11} \over 9}.{{15} \over 4} – {{11} \over 9}.{7 \over 4} \cr & = {{11} \over 9}.\left( {{{15} \over 4} – {7 \over 4} – {5 \over 4}} \right) = {{11} \over 9}.{3 \over 4} = {{11} \over {12}}. \cr & g)\left( { – 3,6 + 5{1 \over 3}:0,4} \right).{{10} \over {73}} – {1 \over 3}:\left( {4,8 – 5{1 \over 2}} \right) = \left( {{{ – 18} \over 5} + {{16} \over 3}:{2 \over 5}} \right).{{10} \over {73}} – {1 \over 3}:\left( {{{24} \over 5} – {{11} \over 2}} \right) \cr & = \left( {{{ – 18} \over 5} + {{16} \over 3}.{5 \over 2}} \right).{{10} \over {73}} – {1 \over 3}:\left( {{{48} \over {10}} – {{55} \over {10}}} \right) = \left( {{{ – 18} \over 5} + {{40} \over 3}} \right).{{10} \over {73}} – {1 \over 3}:{{ – 7} \over {10}} \cr & = \left( {{{ – 54} \over {15}} + {{200} \over {15}}} \right).{{10} \over {73}} – {1 \over 3}.{{10} \over { – 7}} = {{146} \over {15}}.{{10} \over {73}} – {{ – 10} \over {21}} = {4 \over 3} + {{10} \over {21}} = {{28} \over {21}} + {{10} \over {21}} = {{38} \over {21}} = 1{{17} \over {21}}. \cr} \)