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Tính bằng hai cách :
a) \(3{5 \over 9} + 2{1 \over 4}\) b) \(5{1 \over 6} – 3{2 \over 5}\)
c) \(5 – 3{4 \over 7}\) d) \(8{2 \over 7} – \left( {3{4 \over 9} + 4{2 \over 7}} \right)\).
a)Cách 1: \(3{5 \over 9} + 2{1 \over 4} = {{32} \over 9} + {9 \over 4} = {{128} \over {36}} + {{81} \over {36}} = {{209} \over {36}} = 5{{29} \over {36}}.\)
Cách 2: \(3{5 \over 9} + 2{1 \over 4} = 3{{20} \over {36}} + 2{9 \over {36}} = 5{{29} \over {36}}.\)
b) Cách 1: \(5{1 \over 6} – 3{2 \over 5} = {{31} \over 6} – {{17} \over 5} = {{155} \over {30}} – {{102} \over {30}} = {{53} \over {30}} = 1{{23} \over {30}}.\)
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Cách 2: \(5{1 \over 6} – 3{2 \over 5} = 5{5 \over {30}} – 3{{12} \over {30}} = 4{{35} \over {30}} – 3{{12} \over {30}} = 1{{23} \over {30}}.\)
c) Cách 1: \(5 – 3{4 \over 7} = 5 – {{25} \over 7} = {{35} \over 7} – {{25} \over 7} = {{10} \over 7} = 1{3 \over 7}.\)
Cách 2: \(5 – 3{4 \over 7} = 4{7 \over 7} – 3{4 \over 7} = 1{3 \over 7}.\)
d) Cách 1:
\(\eqalign{ & 8{2 \over 7} – \left( {3{4 \over 9} + 4{2 \over 7}} \right) = {{58} \over 7} – \left( {{{31} \over 9} + {{30} \over 7}} \right) \cr & = {{58} \over 7} – {{31} \over 9} – {{30} \over 7} = {{522} \over {63}} – {{217} \over {63}} – {{270} \over {63}} = {{35} \over {63}} = {5 \over 9}. \cr} \)
Cách 2: \(8{2 \over 7} – \left( {3{4 \over 9} + 4{2 \over 7}} \right) = \left( {8{2 \over 7} – 4{2 \over 7}} \right) – 3{4 \over 9} = 4 – 3{4 \over 9} = 3{9 \over 9} – 3{4 \over 9} = {5 \over 9}.\)