Câu 112 trang 31 SBT Toán Lớp 6 tập 2: Tính....

Tính

a) ({ m{}}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) ({ m{}} – 2{1 over 3} – 1{2 over 7})

Giải

a) ({ m{}}6{3 over 8} + 5{1 over 2} = left( {6 + 5} ight) + left( {{3 over 8} + {1 over 2}} ight) )

(= 11 + left( {{3 over 8} + {4 over 8}} ight) = 11 + {7 over 8} = 11{7 over 8}) 

b) (5{3 over 7} – 2{3 over 7} = left( {5 – 2} ight) + left( {{3 over 7} – {3 over 7}} ight) = 3) 

(eqalign{
& c) – 5{1 over 7} + 3{2 over 5} = left( { – 5 + 3} ight) + left( {{{ – 1} over 7} + {2 over 5}} ight) cr
& = – 2 + left( {{{ – 5} over {35}} + {{14} over {35}}} ight) = – 2 + {9 over {35}} cr
& = – 1 – {{35} over {35}} + {9 over {35}} = – 1{{26} over {35}} cr} )

(eqalign{
& { m{d}})- 2{1 over 3} – 1{2 over 7} = – left( {2{1 over 3} + 1{2 over 7}} ight) cr
& = – left[ {left( {2 + 1} ight) + left( {{1 over 3} + {2 over 7}} ight)} ight] cr
& = – left[ {3 + left( {{7 over {21}} + {6 over {21}}} ight)} ight] = – 3{{13} over {21}} cr} )