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