Tính:
a) \(1{{13} \over {15}}.{\left( {0,5} \right)^2}.3 + \left( {{8 \over {15}} - 1{{19} \over {60}}} \right):1{{23} \over {24}}\)
b) \({{\left( {{{{{11}^2}} \over {200}} + 0,415} \right):0,01} \over {{1 \over {12}} - 37,25 + 3{1 \over 6}}}\)
Hướng dẫn làm bài:
a) \(1{{13} \over {15}}.{\left( {0,5} \right)^2}.3 + \left( {{8 \over {15}} - 1{{19} \over {60}}} \right):1{{23} \over {24}}\)
\( = {{28} \over {15}}.{\left( {{1 \over 2}} \right)^2}.3 + \left( {{8 \over {15}} - {{79} \over {60}}} \right):{{47} \over {24}}\)
\( = {{28} \over {15}}.{1 \over 4}.3 + {{8.4 - 79} \over {60}}:{{47} \over {24}}\)
\( = {7 \over 4} + {{ - 47} \over {60}}.{{24} \over {47}}\)
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\( = {7 \over 5} + {{ - 2} \over 5}\)
\( = {5 \over 5} = 1\)
b) \({{\left( {{{{{11}^2}} \over {200}} + 0,415} \right):0,01} \over {{1 \over {12}} - 37,25 + 3{1 \over 6}}}\)
\( = {{\left( {{{121} \over {200}} + {{415} \over {1000}}} \right):{1 \over {100}}} \over {{1 \over {12}} - {{149} \over 4} + {{19} \over 6}}}\)
\( = {{\left( {{{121} \over {200}} + {{83} \over {200}}} \right):{1 \over {100}}} \over {{{1 - 447 + 38} \over {12}}}}\)
\( = \left( {{{204} \over {200}}:{1 \over {100}}} \right):{{ - 408} \over {12}}\)
\( = {{102} \over {100}}.{{100} \over 1}.{{ - 12} \over {408}}\)
\( = {{ - 12} \over 4} = - 3\)