Tính :
\(\eqalign{ & a)\left( { – 3,2} \right).{{ – 15} \over {64}} + \left( {0,8 – 2{4 \over {15}}} \right):3{2 \over 3} \cr & b)\left( { – 6,17 + 3{5 \over 9} – 2{{36} \over {97}}} \right).\left( {{1 \over 3} – 0,25 – {1 \over {12}}} \right). \cr} \)
Advertisements (Quảng cáo)
\(\eqalign{ & a)( – 3,2).{{ – 15} \over {64}} + \left( {0.8 – 2{4 \over {15}}} \right):3{2 \over 3} = {{ – 32} \over {10}}.{{ – 15} \over {64}} + \left( {{8 \over {10}} – {{34} \over {15}}} \right):{{11} \over 3} = {3 \over 4} + \left( {{4 \over 5} – {{34} \over 5}} \right):{{11} \over 3} \cr & = {3 \over 4} + \left( {{{12} \over {15}} – {{34} \over {15}}} \right):{{11} \over 3} = {3 \over 4} + {{ – 22} \over {15}}.{3 \over {11}} = {3 \over 4} + {{ – 2} \over 5} = {{15} \over {20}} + {{ – 8} \over {20}} = {7 \over {20}} \cr & b)\left( { – 6,17 + 3{5 \over 9} – 2{{36} \over {97}}} \right).\left( {{1 \over 3} = 0,25 – {1 \over {12}}} \right) = \left( { – 6,17 + 3{5 \over 9} – 2{{36} \over {97}}} \right).\left( {{1 \over 3} – {1 \over 4} – {1 \over {12}}} \right) \cr & = \left( { – 6,17 + 3{5 \over 9} – 2{{36} \over {97}}} \right).\left( {{4 \over {12}} – {3 \over {12}} – {1 \over {12}}} \right) = \left( { – 6,17 + 3{5 \over 9} – 2{{36} \over {97}}} \right).{0 \over {12}} \cr & = \left( { – 6,17 + 3{5 \over 9} – 2{{36} \over {97}}} \right).0 = 0 \cr} \)