Đề bài

a) \({5 \over {4a}} - {3 \over {8a}}\) ;

b) \({{5b} \over {3b + 2c}} + {b \over {9b + 6c}}\) ;

c) \({9 \over {2d - 4f}} + {7 \over {8f - 4d}}\) ;

d) \({3 \over {k - 1}} - {2 \over {4k + 5}}\) ;

e) \({m \over {2m - 5}} + {4 \over {8 + 3m}}\) ;

f) \({1 \over {{p^2} + 3p - 4}} - {1 \over {p + 4}}\) .

Lời giải chi tiết

\(\eqalign{  & a)\,\,{5 \over {4a}} - {3 \over {8a}} = {{5.2} \over {4a.2}} - {3 \over {8a}} = {{10} \over {8a}} + {{ - 3} \over {8a}} = {7 \over {8a}}  \cr  & b)\,\,{{5b} \over {3b + 2c}} + {b \over {9b + 6c}} = {{5b.3} \over {\left( {3b + 2c} \right).3}} + {b \over {\left( {3b + 2c} \right).3}}  \cr  & \,\,\,\,\, = {{15b + b} \over {\left( {3b + 2c} \right).3}} = {{16b} \over {\left( {3b + 2c} \right).3}}  \cr  & c)\,\,{9 \over {2d - 4f}} + {7 \over {8f - 4d}} = {{9\left( { - 2} \right)} \over {\left( {2d - 4f} \right)\left( { - 2} \right)}} + {7 \over {8f - 4d}}  \cr  & \,\,\,\,\, = {{ - 18} \over {8f - 4d}} + {7 \over {8f - 4d}} = {{ - 11} \over {8f - 4d}}  \cr  & d)\,\,{3 \over {k - 1}} - {2 \over {4k + 5}} = {{3\left( {4k + 5} \right)} \over {\left( {k - 1} \right)\left( {4k + 5} \right)}} + {{ - 2\left( {k - 1} \right)} \over {\left( {4k + 5} \right)\left( {k - 1} \right)}}  \cr  & \,\,\,\,\, = {{12k + 15 - 2k + 2} \over {\left( {k - 1} \right)\left( {4k + 5} \right)}} = {{10k + 17} \over {\left( {k - 1} \right)\left( {4k + 5} \right)}}  \cr  & e)\,\,{m \over {2m - 5}} + {4 \over {8 + 3m}} = {{m\left( {8 + 3m} \right)} \over {\left( {2m - 5} \right)\left( {8 + 3m} \right)}} + {{4\left( {2m - 5} \right)} \over {\left( {8 + 3m} \right)\left( {2m - 5} \right)}}  \cr  & \,\,\,\,\, = {{8m + 3{m^2} + 8m - 20} \over {\left( {2m - 5} \right)\left( {8 + 3m} \right)}} = {{3{m^2} + 16m - 20} \over {\left( {2m - 5} \right)\left( {8 + 3m} \right)}}  \cr  & f)\,\,{1 \over {{p^2} + 3p - 4}} - {1 \over {p + 4}} = {1 \over {\left( {p + 4} \right)\left( {p - 1} \right)}} + {{ - 1} \over {p + 4}}  \cr  & \,\,\,\,\, = {1 \over {\left( {p + 4} \right)\left( {p - 1} \right)}} + {{ - 1\left( {p - 1} \right)} \over {\left( {p + 4} \right)\left( {p - 1} \right)}} = {{1 - p + 1} \over {\left( {p + 4} \right)\left( {p - 1} \right)}} = {{2 - p} \over {\left( {p + 4} \right)\left( {p - 1} \right)}} \cr} \)

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