Bài 91 trang 27 SBT toán 6 tập 2

Đề bài

Áp dụng các tính chất của phép nhân phân số để tính nhanh : 

\(\displaystyle M = {8 \over 3}.{2 \over 5}.{3 \over 8}.10.{{19} \over {92}}\)

\(\displaystyle N = {5 \over 7}.{5 \over {11}} + {5 \over 7}.{2 \over {11}} - {5 \over 7}.{{14} \over {11}}\)

\(\displaystyle Q = \left( {{1 \over {99}} + {{12} \over {999}} - {{123} \over {9999}}} \right)\)\(\displaystyle.\left( {{1 \over 2} - {1 \over 3} - {1 \over 6}} \right)\) 

Lời giải

\(\displaystyle M = {8 \over 3}.{2 \over 5}.{3 \over 8}.10.{{19} \over {92}} \)

\(\displaystyle= \left( {{8 \over 3}.{3 \over 8}} \right).\left( {{2 \over 5}.10} \right).{{19} \over {92}} \)

\(\displaystyle= 1.4.{{19} \over {92}}= {{4.19} \over {92}} \)

\(\displaystyle ={{4.19} \over {4.23}}= {{19} \over {23}}\) 

\(\displaystyle N = {5 \over 7}.{5 \over {11}} + {5 \over 7}.{2 \over {11}} - {5 \over 7}.{{14} \over {11}} \)

\(\displaystyle= {5 \over 7}.\left( {{5 \over {11}} + {2 \over {11}} - {{14} \over {11}}} \right)\)

\(\displaystyle= {5 \over 7}.{{ - 7} \over {11}} = {{ - 5} \over {11}}\) 

\(\displaystyle Q = \left( {{1 \over {99}} + {{12} \over {999}} - {{123} \over {9999}}} \right)\)\(\displaystyle.\left( {{1 \over 2} - {1 \over 3} - {1 \over 6}} \right) \)
\(\displaystyle Q = \left( {{1 \over {99}} + {{12} \over {999}} - {{123} \over {999}}} \right)\)\(\displaystyle.\left( {{3 \over 6} + {{ - 2} \over 6} + {{ - 1} \over 6}} \right) \)
\(\displaystyle Q = \left( {{1 \over {99}} + {{12} \over {999}} - {{123} \over {9999}}} \right).0 = 0  \)


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