Bài 7 trang 57 Tài liệu dạy – học toán 6 tập 2 — Không quảng cáo

Đề bài

Tính :

\(\eqalign{  & a)\left( {6 - 2{4 \over 5}} \right).3{1 \over 8} - 1{3 \over 5}:{1 \over 4}  \cr  & b)14:\left( {4{1 \over {12}} - 2{5 \over 8}} \right) + 14{1 \over 2}.{1 \over 2}  \cr  & c){7 \over 5}.{{15} \over {49}} - \left( {{4 \over 5} + {2 \over 3}} \right):2{1 \over 5}  \cr  & d)\left( {1{1 \over 6}.{6 \over 7} - 6:{3 \over 5}} \right):\left( {4{1 \over 5}.{{10} \over {11}} + 5{2 \over {11}}} \right). \cr} \)

Lời giải chi tiết

\(\eqalign{  & a)\left( {6 - 2{4 \over 5}} \right).3{1 \over 8} - 1{3 \over 5}:{1 \over 4} = \left( {6 - {{14} \over 5}} \right).{{25} \over 8} - {8 \over 5}:{1 \over 4}  \cr  &  = \left( {{{30} \over 5} - {{14} \over 5}} \right).{{25} \over 8} - {8 \over 5}.{4 \over 1} = {{16} \over 5}.{{25} \over 8} - {{32} \over 5} = 10 - {{32} \over 5} = {{50} \over 5} - {{32} \over 5} = {{18} \over 5} = 3{3 \over 5}.  \cr  & b)14:\left( {4{1 \over {12}} - 2{5 \over 8}} \right) + 14{1 \over 2}.{1 \over 2} = 14:\left( {{{49} \over {12}} - {{21} \over 8}} \right) + {{29} \over 2}.{1 \over 2} = 14:\left( {{{98} \over {24}} - {{63} \over {24}}} \right) + {{29} \over 4}  \cr  &  = 14:{{35} \over {24}} + {{29} \over 4} = 14{{24} \over {35}} + {{29} \over 4} = {{48} \over 5} + {{29} \over 4} = {{192} \over {20}} + {{145} \over {20}} = {{337} \over {20}} = 16{{17} \over {20}}  \cr  & c){7 \over 5}.{{15} \over {49}} - \left( {{4 \over 5} + {2 \over 3}} \right):2{1 \over 5} = {3 \over 7} - \left( {{{12} \over {15}} + {{10} \over {15}}} \right):{{11} \over 5}  \cr  &  = {3 \over 7} - {{22} \over {15}}:{{11} \over 5} - {3 \over 7} - {{22} \over {15}}.{5 \over {11}} = {3 \over 7} - {2 \over 3} = {9 \over {21}} - {{14} \over {21}} = {{ - 5} \over {21}}  \cr  & d)\left( {1{1 \over 6}.{6 \over 7} - 6:{3 \over 5}} \right):\left( {4{1 \over 5}.{{10} \over {11}} + 5{2 \over {11}}} \right) = \left( {{7 \over 6}.{6 \over 7} - 6{5 \over 3}} \right):\left( {{{21} \over 5}.{{10} \over {11}} + {{57} \over {11}}} \right)  \cr  &  = (1 - 10):\left( {{{42} \over {11}} + {{57} \over {11}}} \right) =  - 9:{{99} \over {11}} =  - 9:9 =  - 1. \cr} \)