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
\[\eqalign{ & a]{1 \over 3}x - {2 \over 5} = 1{1 \over 3} \cr & {1 \over 3}x - {2 \over 5} = {4 \over 3} \cr & {5 \over {15}}x - {6 \over {15}} = {{20} \over {15}} \cr & 5x - 6 = 20 \cr & 5x = 20 + 6 = 26 \cr & x = {{26} \over 5} \Leftrightarrow x = 5{1 \over 5} \cr & b]{2 \over 7} - {3 \over {14}}x = {1 \over 4} \cr & {8 \over {28}} - {6 \over {28}}x = {7 \over {28}} \cr & 8 - 6x = 7 \cr & 6x = 8 - 7 \cr & 6x = 1 \cr & x = {1 \over 6} \cr & c]4 - \left[ {{1 \over 2}x + {3 \over 4}} \right] = - 1,5 \cr & 4 - \left[ {{1 \over 2}x + {3 \over 4}} \right] = {{ - 3} \over 2} \cr & {1 \over 2}x + {3 \over 4} = 4 + {3 \over 2} \cr & {2 \over 4}x + {3 \over 4} = {{16} \over 4} + {6 \over 4} \cr & 2x + 3 = 16 + 6 = 32 \cr & 2x = 22 - 3 = 19 \cr & x = {{19} \over 2} \Leftrightarrow x = 9{1 \over 2} \cr & d]\left[ {3{1 \over 4} - 2x} \right].2{1 \over 5} = 6{1 \over 5} \cr & \left[ {{{13} \over 4} - 2x} \right].{{11} \over 5} = {{31} \over 5} \cr & {{13} \over 4} - 2x = {{31} \over 5}:{{11} \over 5} \cr & {{13} \over 4} - 2x = {{31} \over {11}} \cr & 2x = {{13} \over 4} - {{31} \over {11}} \cr & 2x = {{143} \over {44}} - {{124} \over {44}} = {{19} \over {44}} \cr & x = {{19} \over {44}}:2 \Leftrightarrow x = {{19} \over {88}}. \cr} \]
\[\eqalign{ & e]{1 \over 2}x + {2 \over 5}.\left[ {2{1 \over 2} - 5x} \right] = 2{1 \over 4} \cr & {1 \over 2}x + {2 \over 5}.\left[ {{5 \over 2} - 5x} \right] = {9 \over 4} \cr & {1 \over 2}x + {2 \over 5}.{5 \over 2} - {2 \over 5}.5x = {9 \over 4} \cr & {1 \over 2}x + 1 - 2x = {9 \over 4} \cr & \left[ {{1 \over 2} - 2} \right]x = {9 \over 4} - 1 \cr & \left[ {{1 \over 2} - {4 \over 2}} \right]x = {9 \over 4} - {4 \over 4} \cr & {{ - 3} \over 2}x = {5 \over 4} \cr & x = {5 \over 4}.{{ - 2} \over 3} \Leftrightarrow x = {{ - 5} \over 6} \cr & f]x:\left[ {9{1 \over 2} - {3 \over 2}} \right] = {{0,4 + {2 \over 9} - {2 \over {11}}} \over {1,6 + {8 \over 9} - {8 \over {11}}}} \cr & x:\left[ {{{19} \over 2} - {3 \over 2}} \right] = {{{2 \over 5} + {2 \over 9} - {2 \over {11}}} \over {{8 \over 5} + {8 \over 9} - {8 \over {11}}}} \cr & x:8 = {{2\left[ {{1 \over 5} + {1 \over 9} - {1 \over {11}}} \right]} \over {8\left[ {{1 \over 5} + {1 \over 9} - {1 \over {11}}} \right]}} \cr & x:8 = {1 \over 4} \cr & x = {1 \over 4}.8 \Leftrightarrow x = 2. \cr} \]