Bài tập cuối chương 1

Lê Nhật Ninh

Tính giá trị các biểu thức sau:

a)A=\(\left[\left(-0,5\right)-\dfrac{2}{3}\right]:\left(-3\right)+\dfrac{1}{3}-\left(-\dfrac{1}{6}\right):\left(2\right)\)

b)B=\(\left(\dfrac{2}{25}-0,036\right):\left(\dfrac{11}{50}\right)-\left[\left(3\dfrac{1}{4}-2\dfrac{4}{9}\right)\right].\dfrac{9}{29}\)

 

Hà Quang Minh
19 tháng 9 2023 lúc 20:50

a)

\(\begin{array}{l}A = \left[ {\left( { - 0,5} \right) - \frac{3}{5}} \right]:\left( { - 3} \right) + \frac{1}{3} - \left( { - \frac{1}{6}} \right):\left( { - 2} \right)\\ = \left( {\frac{{ - 5}}{{10}} - \frac{6}{{10}}} \right).\frac{{ - 1}}{3} + \frac{1}{3} - \frac{-1}{6}.\frac{{ - 1}}{2}\\ = \frac{{ - 11}}{{10}}.\frac{{ - 1}}{3} + \frac{1}{3} - \frac{1}{12}\\ = \frac{{11}}{{30}} + \frac{1}{3} + \frac{{ - 1}}{{12}}\\ = \frac{{22}}{{60}} + \frac{{20}}{{60}} - \frac{{ 5}}{{60}}\\ = \frac{{37}}{{60}}\end{array}\)

b)

\(\begin{array}{l}B = \left( {\frac{2}{{25}} - 0,036} \right):\frac{{11}}{{50}} - \left[ {\left( {3\frac{1}{4} - 2\frac{4}{9}} \right)} \right].\frac{9}{{29}}\\ = \left( {\frac{2}{{25}} - \frac{{36}}{{1000}}} \right).\frac{{50}}{{11}} - \left[ {\left( {\frac{{13}}{4} - \frac{{22}}{9}} \right)} \right].\frac{9}{{29}}\\ = \left( {\frac{{10}}{{125}} - \frac{9}{{250}}} \right).\frac{{50}}{{11}} - \left[ {\left( {\frac{{117}}{{36}} - \frac{{88}}{{36}}} \right)} \right].\frac{9}{{29}}\\ = \frac{{ 11}}{{250}}.\frac{{50}}{{11}} - \frac{{29}}{{36}}.\frac{9}{{29}}\\ = \frac{{ 1}}{{5}} - \frac{1}{4}\\ = \frac{{ 4}}{{20}} - \frac{{5}}{{20}}\\ = \frac{{ - 1}}{{20}}\end{array}\)

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