Tìm x, biết:
\(\eqalign{ & a)\,\,x + {1 \over 5} = {2 \over 3} - \left( { - {1 \over 5}} \right) \cr & b)\,\,{5 \over 7} - x = {1 \over 4} - \left( { - {3 \over 5}} \right) \cr & c)\,\,{2 \over 3} - \left( {{3 \over 4} + x} \right) = \sqrt {{1 \over 9}} \cr} \)
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\(\eqalign{ & a)x + {1 \over 5} = {2 \over 3} - \left( { - {1 \over 5}} \right) \cr & x = {2 \over 3} + {1 \over 5} - {1 \over 5} \cr & x = {2 \over 3} \cr & b){5 \over 7} - x = {1 \over 4} - \left( { - {3 \over 5}} \right) \cr & x = {5 \over 7} - {1 \over 4} - {3 \over 5} \cr & x = {{100} \over {140}} - {{35} \over {140}} - {{84} \over {140}} \cr & x = {{ - 19} \over {140}} \cr & c){2 \over 3} - \left( {{3 \over 4} + x} \right) = \sqrt {{1 \over 9}} \cr & {2 \over 3} - \left( {{3 \over 4} + x} \right) = {1 \over 3} \cr & {3 \over 4} + x = {2 \over 3} - {1 \over 3} \cr & {3 \over 4} + x = {1 \over 3} \cr & x = {1 \over 3} - {3 \over 4} = {{ - 5} \over {12}} \cr} \)