Đề bài - bài 3 trang 83 tài liệu dạy – học toán 6 tập 2

Đề bài

Tìm x, biết :

a) \({1 \over 3}x - {2 \over 5} = 1{1 \over 3}\)

b) \({2 \over 7} - {3 \over {14}}x = {1 \over 4}\)

c) \(4 - \left( {{1 \over 2}x + {3 \over 4}} \right) = - 1,5\)

d) \(\left( {3{1 \over 4} - 2x} \right).2{1 \over 5} = 6{1 \over 5}\)

e) \({1 \over 2}x + {2 \over 5}\left( {2{1 \over 2} - 5x} \right) = 2{1 \over 4}\)

f) \(x:\left( {9{1 \over 2} - {3 \over 2}} \right) = {{0,4 + {2 \over 9} - {2 \over {11}}} \over {1,6 + {8 \over 9} - {8 \over {11}}}}\).

Lời giải chi tiết

\(\eqalign{ & a){1 \over 3}x - {2 \over 5} = 1{1 \over 3} \cr & {1 \over 3}x - {2 \over 5} = {4 \over 3} \cr & {5 \over {15}}x - {6 \over {15}} = {{20} \over {15}} \cr & 5x - 6 = 20 \cr & 5x = 20 + 6 = 26 \cr & x = {{26} \over 5} \Leftrightarrow x = 5{1 \over 5} \cr & b){2 \over 7} - {3 \over {14}}x = {1 \over 4} \cr & {8 \over {28}} - {6 \over {28}}x = {7 \over {28}} \cr & 8 - 6x = 7 \cr & 6x = 8 - 7 \cr & 6x = 1 \cr & x = {1 \over 6} \cr & c)4 - \left( {{1 \over 2}x + {3 \over 4}} \right) = - 1,5 \cr & 4 - \left( {{1 \over 2}x + {3 \over 4}} \right) = {{ - 3} \over 2} \cr & {1 \over 2}x + {3 \over 4} = 4 + {3 \over 2} \cr & {2 \over 4}x + {3 \over 4} = {{16} \over 4} + {6 \over 4} \cr & 2x + 3 = 16 + 6 = 32 \cr & 2x = 22 - 3 = 19 \cr & x = {{19} \over 2} \Leftrightarrow x = 9{1 \over 2} \cr & d)\left( {3{1 \over 4} - 2x} \right).2{1 \over 5} = 6{1 \over 5} \cr & \left( {{{13} \over 4} - 2x} \right).{{11} \over 5} = {{31} \over 5} \cr & {{13} \over 4} - 2x = {{31} \over 5}:{{11} \over 5} \cr & {{13} \over 4} - 2x = {{31} \over {11}} \cr & 2x = {{13} \over 4} - {{31} \over {11}} \cr & 2x = {{143} \over {44}} - {{124} \over {44}} = {{19} \over {44}} \cr & x = {{19} \over {44}}:2 \Leftrightarrow x = {{19} \over {88}}. \cr} \)

\(\eqalign{ & e){1 \over 2}x + {2 \over 5}.\left( {2{1 \over 2} - 5x} \right) = 2{1 \over 4} \cr & {1 \over 2}x + {2 \over 5}.\left( {{5 \over 2} - 5x} \right) = {9 \over 4} \cr & {1 \over 2}x + {2 \over 5}.{5 \over 2} - {2 \over 5}.5x = {9 \over 4} \cr & {1 \over 2}x + 1 - 2x = {9 \over 4} \cr & \left( {{1 \over 2} - 2} \right)x = {9 \over 4} - 1 \cr & \left( {{1 \over 2} - {4 \over 2}} \right)x = {9 \over 4} - {4 \over 4} \cr & {{ - 3} \over 2}x = {5 \over 4} \cr & x = {5 \over 4}.{{ - 2} \over 3} \Leftrightarrow x = {{ - 5} \over 6} \cr & f)x:\left( {9{1 \over 2} - {3 \over 2}} \right) = {{0,4 + {2 \over 9} - {2 \over {11}}} \over {1,6 + {8 \over 9} - {8 \over {11}}}} \cr & x:\left( {{{19} \over 2} - {3 \over 2}} \right) = {{{2 \over 5} + {2 \over 9} - {2 \over {11}}} \over {{8 \over 5} + {8 \over 9} - {8 \over {11}}}} \cr & x:8 = {{2\left( {{1 \over 5} + {1 \over 9} - {1 \over {11}}} \right)} \over {8\left( {{1 \over 5} + {1 \over 9} - {1 \over {11}}} \right)}} \cr & x:8 = {1 \over 4} \cr & x = {1 \over 4}.8 \Leftrightarrow x = 2. \cr} \)