Tìm x, biết :
\(\eqalign{ & a)\left( {x – {5 \over {12}}} \right).{9 \over {29}} = – {6 \over {29}} \cr & b)0,5.x – {2 \over 3}.x = {7 \over {12}} \cr & c)5,2.x + 7{2 \over 5} = 6{3 \over 4} \cr & d)\left( {{3 \over 7}x + 1} \right):\left( { – 4} \right) = {{ – 1} \over {28}}. \cr} \)
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\(\eqalign{ & a)\left( {x – {5 \over {12}}} \right).{9 \over {29}} = – {6 \over {29}} \cr & x – {5 \over {12}} = {{ – 6} \over {29}}:{9 \over {29}} \cr & x – {5 \over {12}} = {{ – 6} \over {29}}.{{29} \over 9} \cr & x – {5 \over {12}} = {{ – 2} \over 3} \cr & x = {{ – 2} \over 3} + {5 \over {12}} \cr & x = {{ – 8} \over {12}} + {5 \over {12}} \cr & x = {{ – 3} \over {12}} \Leftrightarrow x = – {1 \over 4}. \cr & b)0,5x – {2 \over 3}x = {7 \over {12}} \Leftrightarrow {5 \over {10}}x – {2 \over 3}x = {7 \over {12}} \cr & \left( {{5 \over {10}} – {2 \over 3}} \right).x = {7 \over {12}} \cr & \left( {{1 \over 2} – {2 \over 3}} \right).x = {7 \over {12}} \cr & \left( {{3 \over 6} – {4 \over 6}} \right).x = {7 \over {12}} \cr & – {1 \over 6}.x = {7 \over {12}} \cr & x = {7 \over {12}}:{{ – 1} \over 6} \Leftrightarrow x = {7 \over {12}}.( – 6) \cr & x = – {7 \over 2} \Leftrightarrow x = – 3{1 \over 2}. \cr & c)5,2x + 7{2 \over 5} = 6{3 \over 4} \Leftrightarrow {{52} \over {10}}x + {{37} \over 5} = {{27} \over 4} \cr & {{26} \over 5}x = {{27} \over 4} – {{37} \over 5} \cr & {{26} \over 5}x = {{135} \over {20}} – {{148} \over {20}} \cr & {{26} \over 5}x = {{ – 13} \over {20}} \cr & x = {{ – 13} \over {20}}:{{26} \over 5} \cr & x = {{ – 13} \over {20}}.{5 \over {26}} \Leftrightarrow x = – {1 \over 8}. \cr & d)\left( {{3 \over 7}x + 1} \right):( – 4) = {{ – 1} \over {28}} \cr & {3 \over 7}x + 1 = {{ – 1} \over {28}}.( – 4) \cr & {3 \over 7}x + 1 = {1 \over 7} \Leftrightarrow {3 \over 7}x = {1 \over 7} – 1 \cr & {3 \over 7}x = {1 \over 7} – {7 \over 7} \Leftrightarrow {3 \over 7}x = {{ – 6} \over 7} \cr & x = {{ – 6} \over 7}:{3 \over 7} \cr & x = {{ – 6} \over 7}.{7 \over 3} \Leftrightarrow x = – 2. \cr} \)