Bài 9 trang 14 Vở bài tập toán 7 tập 1

Đề bài

Tính

\(\eqalign{
& a)\,\,{{ - 3} \over 4}.{{12} \over { - 5}}.\left( {{{ - 25} \over 6}} \right) \cr 
& b)\,\,( - 2).{{ - 38} \over {21}}.{{ - 7} \over 4}.\left( { - {3 \over 8}} \right) \cr 
& c)\,\,\left( {{{11} \over {12}}:{{33} \over {16}}} \right).{3 \over 5} \cr 
& d)\,\,{7 \over {23}}.\left[ {\left( { - {8 \over 6}} \right) - {{45} \over {18}}} \right] \cr} \)

Lời giải chi tiết

\(\eqalign{
& a)\,\,{{ - 3} \over 4}.{{12} \over { - 5}}.\left( {{{ - 25} \over 6}} \right) \cr 
& = {{ - 3} \over 4}.{{ - 12} \over 5}.{{ - 25} \over 6} \cr 
& = {{\left( { - 3} \right).( - 12).( - 25)} \over {4.5.6}} \cr 
& = {{\left( { - 3} \right).\left( { - 2} \right).6.\left( { - 5} \right).5} \over {2.2.5.6}} \cr 
& = {{ - 3.5} \over 2} = - {{15} \over 2} \cr 
& b)\,\,( - 2).{{ - 38} \over {21}}.{{ - 7} \over 4}.\left( { - {3 \over 8}} \right)\cr&=\frac{{ - 2}}{1}.\frac{{ - 38}}{{21}}.\frac{{ - 7}}{4}.\frac{{ - 3}}{8} \cr 
& = {{( - 2).( - 38).( - 7).( - 3)} \over {21.4.8}} \cr 
& = {{38} \over {2.8}} = {{19} \over 8} = 2{3 \over 8} \cr 
& c)\,\,\left( {{{11} \over {12}}:{{33} \over {16}}} \right).{3 \over 5} \cr 
& = \left( {{{11} \over {12}}.{{16} \over {33}}} \right).{3 \over 5} \cr&= \frac{{11.16.3}}{{12.33.5}} = \frac{4}{{3.5}} = \frac{4}{{15}} \cr 
& d)\,\,{7 \over {23}}.\left[ {\left( { - {8 \over 6}} \right) - {{45} \over {18}}} \right] \cr 
& = \,{7 \over {23}}.\left[ {{\frac{{ - 8}}{6}} - {{15} \over 6}} \right]  \cr 
& = {7 \over {23}}.{{ - 23} \over 6} = {{ - 7} \over 6} = - 1{1 \over 6} \cr} \)