Tính
\(E = {{\left( {13{1 \over 4} – 2{5 \over {27}} – 10{5 \over 6}} \right).230{1 \over {25}} + 46{3 \over 4}} \over {\left( {1{3 \over 7} + {{10} \over 3}} \right):\left( {12{1 \over 3} – 14{2 \over 7}} \right)}}\)
Giải
\(E = {{\left( {13{1 \over 4} – 2{5 \over {27}} – 10{5 \over 6}} \right).230{1 \over {25}} + 46{3 \over 4}} \over {\left( {1{3 \over 7} + {{10} \over 3}} \right):\left( {12{1 \over 3} – 14{2 \over 7}} \right)}}\)
\(\eqalign{
& = {{\left( {13 – 2 – 10 + {1 \over 4} – {5 \over {27}} – {5 \over 6}} \right).{{5771} \over {25}} + {{187} \over 4}} \over {\left( {{{30} \over {21}} + {{70} \over {21}}} \right):\left( {{{259} \over {21}} – {{300} \over {21}}} \right)}} \cr
& = {{\left( {1 + {{27 – 20 – 90} \over {108}}} \right).{{5751} \over {25}} + {{187} \over 4}} \over {{{100} \over {21}}:{{ – 41} \over {21}}}} \cr} \)
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\(\eqalign{
& = {{\left( {{{108} \over {108}} – {{83} \over {108}}} \right).{{5751} \over {25}} + {{187} \over 4}} \over {{{100} \over {21}}.{{ – 21} \over {41}}}} \cr
& = {{{{25} \over {108}}.{{5751} \over {25}} + {{187} \over 4}} \over {{{ – 100} \over {41}}}} \cr
& = \left( {{{213} \over 4} + {{187} \over 4}} \right).{{ – 41} \over {100}} = 100.{{ – 41} \over {100}} \cr} \)
= -41
Mục lục môn Toán 7 (SBT)