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a) \({\rm{4}}{3 \over 4} + \left( { – 0,37} \right) + {1 \over 8} + \left( { – 1,28} \right) + \left( { – 2,5} \right) + 3{1 \over {12}}\)
b) \({3 \over {5.7}} + {3 \over {7.9}} + .. + {3 \over {59.61}}\)
c) \({{{5 \over {22}} + {3 \over {13}} – {1 \over 2}} \over {{4 \over {13}} – {2 \over {11}} + {3 \over 2}}}\)
Giải
a) \({\rm{4}}{3 \over 4} + \left( { – 0,37} \right) + {1 \over 8} + \left( { – 1,28} \right) + \left( { – 2,5} \right) + 3{1 \over {12}}\)
\(\eqalign{
& = \left( {4{3 \over 9} + {1 \over 8} + 3{1 \over {12}}} \right) – \left( {0,37 + 1,28 + 2,5} \right) \cr
& = \left( {4{{18} \over {24}} + {3 \over {24}} + 3{2 \over {24}}} \right) – \left( {4,15} \right) \cr
& = 7{{23} \over {24}} – 4{3 \over {20}} = 7{{115} \over {120}} – 4{{18} \over {120}} = 3{{97} \over {120}} \cr} \)
\(\eqalign{
& b){3 \over {5.7}} + {3 \over {7.9}} + .. + {3 \over {59.61}} \cr
& = {3 \over 2}.\left( {{2 \over {5.7}} + {2 \over {7.9}} + .. + {2 \over {59.61}}} \right) \cr
& = {3 \over 2}.\left( {{1 \over 5} – {1 \over 7} + {1 \over 7} – {1 \over 9} + … + {1 \over {59}} – {1 \over {61}}} \right) \cr
& = {3 \over 2}.\left( {{1 \over 5} – {1 \over {61}}} \right) \cr
& = {3 \over 2}.{{56} \over {305}} = {{84} \over {305}} \cr} \)
\(\eqalign{
& c){{{5 \over {22}} + {3 \over {13}} – {1 \over 12}} \over {{4 \over {13}} – {2 \over {11}} + {3 \over 2}}} = {{\left( {{5 \over {22}} + {3 \over {13}} – {1 \over 12}} \right).\left( {2.11.13} \right)} \over {\left( {{4 \over {13}} – {2 \over {11}} + {3 \over 2}} \right).\left( {2.11.13} \right)}} \cr
& = {{65 + 66 – 143} \over {88 – 52 + 429}} = {{ – 12} \over {465}} = {{ – 4} \over {155}} \cr} \)