Bài 90 trang 27 SBT toán 6 tập 2

Đề bài

Tính nhanh giá trị các biểu thức sau:

\(\displaystyle{\rm{A}} = {6 \over 7} + {1 \over 7}.{2 \over 7} + {1 \over 7}.{5 \over 7};\)                        \(\displaystyle B = {4 \over 9}.{{13} \over 3} - {4 \over 3}.{{40} \over 9}.\)

Lời giải chi tiết

Ta có: 

\(\begin{array}{l}
A = \dfrac{6}{7} + \dfrac{1}{7}.\dfrac{2}{7} + \dfrac{1}{7}.\dfrac{5}{7}\\
= \dfrac{6}{7} + \dfrac{1}{7}.\left( {\dfrac{2}{7} + \dfrac{5}{7}} \right)\\
= \dfrac{6}{7} + \dfrac{1}{7}.\dfrac{7}{7}\\
= \dfrac{6}{7} + \dfrac{1}{7} = \dfrac{{6 + 1}}{7} = \dfrac{7}{7} = 1\\
B = \dfrac{4}{9}.\dfrac{{13}}{3} - \dfrac{4}{3}.\dfrac{{40}}{9} = \dfrac{{4.13}}{{9.3}} - \dfrac{{4.40}}{{9.3}}\\
= \dfrac{{4.13 - 4.40}}{{9.3}} = \dfrac{{4.\left( {13 - 40} \right)}}{{9.3}}\\
= \dfrac{{4.\left( { - 27} \right)}}{{27}} = - 4
\end{array}\)

Cách khác: 

\(\displaystyle{\rm{A}} = {6 \over 7} + {1 \over 7}.{2 \over 7} + {1 \over 7}.{5 \over 7} \)

\(\displaystyle= {1 \over 7}.\left( {6 + {2 \over 7} + {5 \over 7}} \right) \)

\(\displaystyle= {1 \over 7}.7 = 1\) 

\(\displaystyle B = {4 \over 9}.{{13} \over 3} - {4 \over 3}.{{40} \over 9} \)

\(\displaystyle B = {4 \over 9}.{{13} \over 3} - {4 \over 9}.{{40} \over 3} \)

\(\displaystyle = {4 \over 9}.\left( {{{13} \over 3} - {{40} \over 3}} \right) \)

\(\displaystyle= {4 \over 9}.{{ - 27} \over 3} = {{4.( - 27)} \over {27}} =  - 4\) 

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