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