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