Câu 139 trang 34 Sách Bài Tập (SBT) Toán lớp 7 tập 1

Tính

\(G = {{4,5:\left[ {47,375 - \left( {26{1 \over 3} - 18.0,75} \right).2,4:0,88} \right]} \over {17,81:1,37 - 23{2 \over 3}:1{5 \over 6}}}\)

Giải

\(G = {{4,5:\left[ {47,375 - \left( {26{1 \over 3} - 18.0,75} \right).2,4:0,88} \right]} \over {17,81:1,37 - 23{2 \over 3}:1{5 \over 6}}}\) 

\( = {{4,5:\left[ {47,375 - \left( {{{79} \over 3} - 18.{3 \over 4}} \right).2{2 \over 5}:{{22} \over {25}}} \right]} \over {13 - {{71} \over 3}:{{11} \over 6}}}\) 

\(\eqalign{
& = {{4,5:\left[ {47{3 \over 8} - \left( {{{158} \over 6} - {{81} \over 6}} \right).{{12} \over 5}:{{22} \over {25}}} \right]} \over {13 - {{142} \over {11}}}} \cr
& = {{4,5:\left[ {47{3 \over 8} - {{77} \over 6}.{{12} \over 5}:{{22} \over {25}}} \right]} \over {{{143} \over {11}} - {{142} \over {11}}}} \cr} \)

\(\eqalign{
& = {{4,5:\left[ {47{3 \over 8} - {{154} \over 5}.{{25} \over {22}}} \right]} \over {{1 \over {11}}}} \cr
& = {{4,5:\left[ {47{3 \over 8} - 35} \right]} \over {{1 \over {11}}}} \cr} \)

\(\eqalign{
& = \left( {4,5:12{3 \over 8}} \right):{1 \over {11}} = 4,5.{8 \over {99}}.{{11} \over 1} \cr
& = {{4,5.8.11} \over {99}} = 4 \cr} \)

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