Bài 95 trang 45 SGK Toán 7 tập 1

Tính giá trị biểu thức: \(\eqalign{
& A = - 5,13:\left( {5{5 \over {28}} - 1{8 \over 9}.1,25 + 1{{16} \over {63}}} \right) \cr
& B = \left( {3{1 \over 3}.1,9 + 19,5:4{1 \over 3}} \right).\left( {{{62} \over {75}} - {4 \over {25}}} \right) \cr}\)

Lời giải

\(\eqalign{
& A = - 5,13:\left( {5{5 \over {28}} - 1{8 \over 9}.1,25 + 1{{16} \over {63}}} \right) \cr
& = - 5,13:\left( {{{145} \over {28}} - {{17} \over 9}.{{125} \over {100}} + {{79} \over {63}}} \right) \cr
& = - 5,13:\left( {{{145} \over {28}} - {{17} \over 9}.{5 \over 4} + {{79} \over {63}}} \right) \cr
& = - 5,13:\left( {{{145} \over {28}} - {{85} \over {36}} + {{79} \over {63}}} \right)\cr& =  - 5,13:\left( {{{1305} \over {252}} - {{595} \over {252}} + {{316} \over {252}}} \right) \cr
& =  - 5,13:{{1026} \over {252}} = {{ - 513} \over {100}}:{{57} \over {14}} \cr
& = {{ - 513} \over {100}}.{{14} \over {57}} = {{ - 9.7} \over {50.1}} = {{ - 63} \over {50}} \cr} \)

\(\eqalign{
& B = \left( {3{1 \over 3}.1,9 + 19,5:4{1 \over 3}} \right).\left( {{{62} \over {75}} - {4 \over {25}}} \right) \cr
& = \left( {{{10} \over 3}.{{19} \over {10}} + {{195} \over {10}}:{{13} \over 3}} \right).\left( {{{62} \over {75}} - {{12} \over {75}}} \right) \cr
& = \left( {{{19} \over 3} + {{39} \over 2}.{3 \over {13}}} \right).{{50} \over {75}} = \left( {{{19} \over 3} + {{3.3} \over {2.1}}} \right).{2 \over 3} \cr
& = \left( {{{19} \over 3} + {9 \over 2}} \right).{2 \over 3} = \left( {{{38} \over 6} + {{27} \over 6}} \right).{2 \over 3} \cr
& = {{65} \over 6}.{2 \over 3} = {{65.1} \over {3.3}} = {{65} \over 9} \cr} \)


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