Bài 152 trang 40 SBT toán 6 tập 2

Tính :\(\displaystyle 1{{13} \over {15}}.0,75 - \left( {{{104} \over {195}} + 25\% } \right).{{24} \over {47}} \)\(\displaystyle - 3{{12} \over {13}}:3 \)

Lời giải

\(\displaystyle 1{{13} \over {15}}.0,75 - \left( {{{104} \over {195}} + 25\% } \right).{{24} \over {47}}\)\(\displaystyle  - 3{{12} \over {13}}:3\) 

\(\displaystyle = {{28} \over {15}}.{3 \over 4} - \left( {{8 \over {15}} + {1 \over 4}} \right).{{24} \over {47}} \)\(\displaystyle - {{51} \over {13}}.{1 \over 3} \)
\(\displaystyle  = {{28} \over {15}}.{3 \over 4} - \left( {{{32} \over {60}} + {{16} \over {60}}} \right).{{24} \over {47}} \)\(\displaystyle - {{51} \over {13}}.{1 \over 3} \) 
\(\displaystyle  = {7 \over 5} - {{47} \over {60}}.{{24} \over {47}} - {{17} \over {13}} \)

\(\displaystyle  = {7 \over 5} - {2 \over 5} - \left( {1 + \frac{4}{{13}}} \right)\)

\(\displaystyle  = {7 \over 5} - {2 \over 5} - 1 - {4 \over {13}} \)
\(\displaystyle  = 1 - 1 - {4 \over {13}} \)
\(\displaystyle  = - {4 \over {13}} \)


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