Đề bài
Tìm x, biết :
a] \[x.{1 \over 2} = {3 \over 8}\] b] \[x:{2 \over {11}} = 3{2 \over 3}\]
c] \[5{4 \over 7}:x = 13\] d] \[{5 \over 6}.x = {{ - 1} \over 6}:{2 \over 3}\]
e] \[{{ - 2} \over 5} + {2 \over 7}:x = {1 \over 4}\]
f] \[\left[ {x - {2 \over {15}}} \right].{9 \over {25}} = - {6 \over {25}}\].
Lời giải chi tiết
\[\eqalign{ & a]x.{1 \over 2} = {3 \over 8} \cr & x = {3 \over 8}:{1 \over 2} \cr & x = {3 \over 8}.{2 \over 1} \cr & x = {{3.2} \over {8.1}} \Leftrightarrow x = {3 \over 4} \cr & b]x:{2 \over {11}} = {{11} \over 3} \cr & x = {{11} \over 3}.{2 \over {11}} \cr & x = {{11.2} \over {3.11}} \Leftrightarrow x = {2 \over 3} \cr & c]5{4 \over 7}:x = 13 \cr & \Leftrightarrow {{39} \over 7}:x = 13 \cr & x = {{39} \over 7}:13 \cr & x = {{39} \over 7}.{1 \over {13}} \cr & x = {{39.1} \over {7.13}} \Leftrightarrow x = {3 \over 7} \cr & d]{5 \over 6}.x = {{ - 1} \over 6}:{2 \over 3} \cr & {5 \over 6}x = {{ - 1} \over 6}.{3 \over 2} \cr & {5 \over 6}x = {{ - 1.3} \over {6.2}} \cr & {5 \over 6}x = {{ - 1} \over 4} \cr & x = {{ - 1} \over 4}.{6 \over 5} \cr & x = {{[ - 1].6} \over {4.5}} \Leftrightarrow x = {{ - 3} \over {10}} \cr & e]{{ - 2} \over 5} + {2 \over 7}:x = {1 \over 4} \cr & {2 \over 7}:x = {1 \over 4} + {2 \over 5} \cr & {2 \over 7}:x = {5 \over {20}} + {8 \over {20}} \cr & {2 \over 7}:x = {{13} \over {20}} \cr & x = {2 \over 7}:{{13} \over {20}} \cr & x = {2 \over 7}.{{20} \over {13}} \Leftrightarrow x = {{40} \over {91}} \cr & f]\left[ {x - {2 \over {15}}} \right].{9 \over {25}} = {{ - 6} \over {25}} \cr & x - {2 \over {15}} = {{ - 6} \over {25}}:{9 \over {25}} \cr & x - {2 \over {15}} = {{ - 6} \over {25}}.{{25} \over 9} \cr & x - {2 \over {15}} = {{ - 6.25} \over {25.9}} \cr & x - {2 \over {15}} = {{ - 2} \over 3} \cr & x = {{ - 2} \over 3} + {2 \over {15}} \cr & x = {{ - 10} \over {15}} + {2 \over {15}} \Leftrightarrow x = {{ - 8} \over {15}}. \cr} \]