Bài 176 trang 67 SGK Toán 6 tập 2

Đề bài

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 giải

a) $1{{13} \over {15}}.{\left( {0,5} \right)^2}.3 + \left( {{8 \over {15}} - 1{{19} \over {60}}} \right):1{{23} \over {24}}$

$= {{28} \over {15}}.{\left( {{1 \over 2}} \right)^2}.3 + \left( {{8 \over {15}} - {{79} \over {60}}} \right):{{47} \over {24}}$

$= {{28} \over {15}}.{1 \over 4}.3 + {{8.4 - 79} \over {60}}:{{47} \over {24}}$

$= {7 \over 5} + {{ - 47} \over {60}}.{{24} \over {47}}$

$= {7 \over 5} + {{ - 2} \over 5}$

$= {5 \over 5} = 1$

b) ${{\left( {{{{{11}^2}} \over {200}} + 0,415} \right):0,01} \over {{1 \over {12}} - 37,25 + 3{1 \over 6}}}$

$= {{\left( {{{121} \over {200}} + {{415} \over {1000}}} \right):{1 \over {100}}} \over {{1 \over {12}} - {{149} \over 4} + {{19} \over 6}}}$

$= {{\left( {{{121} \over {200}} + {{83} \over {200}}} \right):{1 \over {100}}} \over {{{1 - 447 + 38} \over {12}}}}$

$= \left( {{{204} \over {200}}:{1 \over {100}}} \right):{{ - 408} \over {12}}$

$= {{102} \over {100}}.{{100} \over 1}.{{ - 12} \over {408}}$

$= {{ - 12} \over 4} =  - 3$