Advertisements (Quảng cáo)
Tìm x, biết :
a) \({4 \over 5} + x = {2 \over 3}\)
b) \(x – {5 \over 9} = 1{{ – 2} \over 3}\)
c) \({3 \over 4} – x = {1 \over 3}\)
d) \(1{1 \over 2} – {2 \over 3}x = {7 \over {12}}\)
e) \(2,8:\left( {{1 \over 5} – 3x} \right) = 1{2 \over 5}\)
f)\(\left( {0,25 – {2 \over 3}x} \right):\left( {{4 \over 9}} \right) = {3 \over 2}\).
Advertisements (Quảng cáo)
\(\eqalign{ & a){4 \over 5} + x = {2 \over 3} \cr & x = {2 \over 3} – {4 \over 5} \cr & x = {{10} \over {15}} – {{12} \over {15}} \Leftrightarrow x = {{ – 2} \over {15}} \cr & b)x – {5 \over 9} = 1{{ – 2} \over 3} \cr & x = {{ – 5} \over 3} + {5 \over 9} \cr & x = {{ – 15} \over 9} + {5 \over 9} \cr & x = {{ – 10} \over 9} \Leftrightarrow x = – 1{1 \over 9} \cr & c){3 \over 4} – x = {1 \over 3} \cr & x = {3 \over 4} – {1 \over 3} \cr & x = {9 \over {12}} – {4 \over {12}} \cr & x = {5 \over {12}} \Leftrightarrow x = {{ – 5} \over 8} \cr & \cr} \)
\(\eqalign{ & d)1{1 \over 2} – {2 \over 3}x = {7 \over {12}} \cr & {2 \over 3}x = 1{1 \over 2} – {7 \over 2} \cr & {2 \over 3}x = {3 \over 2} – {7 \over {12}} \cr & {2 \over 3}x = {{18} \over {12}} – {7 \over {12}} \cr & {2 \over 3}x = {{11} \over {12}} \cr & x = {{11} \over {12}}:{2 \over 3} \cr & x = {{11} \over {12}}.{3 \over 2} \Leftrightarrow x = {{11} \over 8} = 1{3 \over 8} \cr & e)2,8:\left( {{1 \over 5} – 3x} \right) = 1{2 \over 5} \cr & {{14} \over 5}:\left( {{1 \over 5} – 3x} \right) = {7 \over 5} \cr & {1 \over 5} – 3x = {{14} \over 5}:{7 \over 5} \cr & {1 \over 5} – 3x = {{14} \over 5}.{5 \over 7} \cr & {1 \over 5} – 3x = 2 \cr & 3x = {1 \over 5} – 2 \cr & 3x = {1 \over 5} – {{10} \over 5} \Leftrightarrow 3x = {{ – 9} \over 5} \cr & x = {{ – 9} \over 5}:3 \Leftrightarrow x = {{ – 3} \over 5} \cr & f)\left( {0,25 – {2 \over 3}x} \right):{4 \over 9} = {3 \over 2} \cr & 0,25 – {2 \over 3}x = {3 \over 2}.{4 \over 9} \cr & {1 \over 4} – {2 \over 3}x = {2 \over 3} \cr & {2 \over 3}x = {{ – 2} \over 3} + {1 \over 4} \cr & {2 \over 3}x = {{ – 8} \over {12}} + {3 \over {12}} \cr & {2 \over 3}x = {{ – 5} \over {12}} \cr & x = {{ – 5} \over {12}}.{3 \over 2} \Leftrightarrow x = {{ – 15} \over {24}}. \cr} \)