Nội dung bài giảng
Tính
a) \({\rm{}}6{3 \over 8} + 5{1 \over 2}\) b) \(5{3 \over 7} - 2{3 \over 7}\)
c) \( - 5{1 \over 7} + 3{2 \over 5}\) d) \({\rm{}} - 2{1 \over 3} - 1{2 \over 7}\)
Giải
a) \({\rm{}}6{3 \over 8} + 5{1 \over 2} = \left( {6 + 5} \right) + \left( {{3 \over 8} + {1 \over 2}} \right) \)
\(= 11 + \left( {{3 \over 8} + {4 \over 8}} \right) = 11 + {7 \over 8} = 11{7 \over 8}\)
b) \(5{3 \over 7} - 2{3 \over 7} = \left( {5 - 2} \right) + \left( {{3 \over 7} - {3 \over 7}} \right) = 3\)
\(\eqalign{
& c) - 5{1 \over 7} + 3{2 \over 5} = \left( { - 5 + 3} \right) + \left( {{{ - 1} \over 7} + {2 \over 5}} \right) \cr
& = - 2 + \left( {{{ - 5} \over {35}} + {{14} \over {35}}} \right) = - 2 + {9 \over {35}} \cr
& = - 1 - {{35} \over {35}} + {9 \over {35}} = - 1{{26} \over {35}} \cr} \)
\(\eqalign{
& {\rm{d}})- 2{1 \over 3} - 1{2 \over 7} = - \left( {2{1 \over 3} + 1{2 \over 7}} \right) \cr
& = - \left[ {\left( {2 + 1} \right) + \left( {{1 \over 3} + {2 \over 7}} \right)} \right] \cr
& = - \left[ {3 + \left( {{7 \over {21}} + {6 \over {21}}} \right)} \right] = - 3{{13} \over {21}} \cr} \)