Bài 152: trang 40 sbt Toán 6 tập 2
Tính
$1{{13} \over {15}}.0,75 - \left( {{{104} \over {195}} + 25\% } \right).{{24} \over {47}} - 3{{12} \over {13}} \div 3$
Bài Làm:
\(1{{13} \over {15}}.0,75 - \left( {{{104} \over {195}} + 25\% } \right).{{24} \over {47}} - 3{{12} \over {13}} \div 3\)
\(= {{28} \over {15}}.{3 \over 4} - \left( {{8 \over {15}} + {1 \over 4}} \right).{{24} \over {47}} - {{51} \over {13}}.{1 \over 3} \)
\(= {7 \over 5}- \left( {{{32} \over {60}} + {{15} \over {60}}} \right).{{24} \over {47}} - {{17} \over {13}}\)
\(= {7 \over 5} - {{47} \over {60}}.{{24} \over {47}} - {{17} \over {13}} \)
\(= {7 \over 5} - {2 \over 5} - {{17} \over {13}} \)
\(= 1 - 1 - {4 \over {13}} \)
\(= - {4 \over {13}} \)