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a) \({1 \over {2b}} – {3 \over {b + c}}\) ;
b) \({2 \over {x + 5}} – {3 \over {x – 1}}\) ;
c) \(x – {{3x} \over {x + 3}}\) ;
d) \({3 \over {4{m^2} – 1}} – {5 \over {2m + 1}}\) .
\(\eqalign{ & a)\,\,{1 \over {2b}} – {3 \over {b + c}} = {1 \over {2b}} + {{ – 3} \over {b + c}} = {{1.\left( {b + c} \right)} \over {2b\left( {b + c} \right)}} + {{ – 3.2b} \over {2b\left( {b + c} \right)}} \cr & \,\,\,\,\, = {{b + c – 6b} \over {2b\left( {b + c} \right)}} = {{ – 5b + c} \over {2b\left( {b + c} \right)}} \cr & b)\,\,{2 \over {x + 5}} – {3 \over {x – 1}} = {2 \over {x + 5}} + {{ – 3} \over {x – 1}} \cr & \,\,\,\,\, = {{2\left( {x – 1} \right)} \over {\left( {x + 5} \right)\left( {x – 1} \right)}} + {{ – 3\left( {x + 5} \right)} \over {\left( {x – 1} \right)\left( {x + 5} \right)}} \cr & \,\,\,\,\, = {{2x – 2 – 3x – 15} \over {\left( {x + 5} \right)\left( {x – 1} \right)}} = {{ – x – 17} \over {\left( {x + 5} \right)\left( {x – 1} \right)}} \cr & c)\,\,x – {{3x} \over {x + 3}} = x + {{ – 3x} \over {x + 3}} = {{x\left( {x + 3} \right) – 3x} \over {x + 3}} \cr & \,\,\,\,\, = {{{x^2} + 3x – 3x} \over {x + 3}} = {{{x^2}} \over {x + 3}} \cr & d)\,\,{3 \over {4{m^2} – 1}} – {5 \over {2m + 1}} = {3 \over {4{m^2} – 1}} + {{ – 5} \over {2m + 1}} \cr & \,\,\,\,\, = {3 \over {\left( {2m – 1} \right)\left( {2m + 1} \right)}} + {{ – 5\left( {2m – 1} \right)} \over {\left( {2m – 1} \right)\left( {2m + 1} \right)}} \cr & \,\,\,\,\, = {{3 – 10m + 5} \over {4{m^2} – 1}} = {{8 – 10m} \over {4{m^2} – 1}} \cr} \)