Bài 176 trang 67 SGK Toán 6 tập 2

Đề bài

Tính:

a) \( \displaystyle 1{{13} \over {15}}.{\left( {0,5} \right)^2}.3 + \left( {{8 \over {15}} - 1{{19} \over {60}}} \right):1{{23} \over {24}}\)

b) \(\dfrac{{\left( {\dfrac{{{{11}^2}}}{{200}} + 0,415} \right):0,01}}{{\dfrac{1}{{12}} - 37,25 + 3\dfrac{1}{6}}}\)

Lời giải chi tiết

a) \( \displaystyle 1{{13} \over {15}}.{\left( {0,5} \right)^2}.3 + \left( {{8 \over {15}} - 1{{19} \over {60}}} \right):1{{23} \over {24}}\)

\( \displaystyle = {{28} \over {15}}.{\left( {{1 \over 2}} \right)^2}.3 + \left( {{8 \over {15}} - {{79} \over {60}}} \right):{{47} \over {24}}\) 

\( \displaystyle = {{28} \over {15}}.{1 \over 4}.3 + {{8.4 - 79} \over {60}}:{{47} \over {24}}\)

\( \displaystyle = {7 \over 5} + {{ - 47} \over {60}}.{{24} \over {47}}\)

\( \displaystyle = {7 \over 5} + {{ - 2} \over 5}\)

\( \displaystyle = {5 \over 5} = 1\)

b) \( \displaystyle \dfrac{{\left( {\dfrac{{{{11}^2}}}{{200}} + 0,415} \right):0,01}}{{\dfrac{1}{{12}} - 37,25 + 3\dfrac{1}{6}}}\)

\( = \dfrac{{\left( {\dfrac{{121}}{{200}} + \dfrac{{415}}{{1000}}} \right):\dfrac{1}{{100}}}}{{\dfrac{1}{{12}} - \dfrac{{149}}{4} + \dfrac{{19}}{6}}}\)

\( = \dfrac{{\left( {\dfrac{{121}}{{200}} + \dfrac{{83}}{{200}}} \right):\dfrac{1}{{100}}}}{{\dfrac{{1 - 447 + 38}}{{12}}}}\)

\(\begin{array}{l}
= \dfrac{{\dfrac{{204}}{{200}}:\dfrac{1}{{100}}}}{{\dfrac{{ - 408}}{{12}}}}\\
= \dfrac{{\dfrac{{102}}{{100}}.100}}{{ - 34}}\\
= \dfrac{{102}}{{ - 34}} = - 3
\end{array}\)