Advertisements (Quảng cáo)
Thực hiện phân thức:
a) \({5 \over {3x}} – {2 \over {3x}}\) ;
b) \({{ – 5} \over {7y}} – {6 \over {7y}}\) ;
c) \({{7x} \over {x – y}} – {{2x + 3} \over {x – y}}\) ;
d) \({1 \over {3c}} – {1 \over {3d}}\) ;
e) \({{f – 4h} \over {3k}} – {{2f – 5h} \over {8k}}\) ;
Advertisements (Quảng cáo)
f) \({{p + 3} \over {2z}} + {{p – 1} \over {6z}} – {{2p + 1} \over {3z}}\) .
\(\eqalign{ & a)\,\,{5 \over {3x}} – {2 \over {3x}} = {{5 – 2} \over {3x}} = {3 \over {3x}} = {1 \over x} \cr & b)\,\,{{ – 5} \over {7y}} – {6 \over {7y}} = {{ – 5} \over {7y}} + {{ – 6} \over {7y}} = {{ – 5 – 6} \over {7y}} = {{ – 11} \over {7y}} \cr & c)\,\,{{7x} \over {x – y}} – {{2x + 3} \over {x – y}} = {{7x} \over {x – y}} + {{ – \left( {2x + 3} \right)} \over {x – y}} = {{7x – 2x – 3} \over {x – y}} = {{5x – 3} \over {x – y}} \cr & d)\,\,{1 \over {3c}} – {1 \over {3d}} = {1 \over {3c}} + {{ – 1} \over {3d}} = {{1.d} \over {3c.d}} + {{ – 1.c} \over {3d.c}} = {{d – c} \over {3cd}} \cr & e)\,\,{{f – 4h} \over {3k}} – {{2f – 5h} \over {8k}} = {{f – 4h} \over {3k}} + {{ – \left( {2f – 5h} \right)} \over {8k}} \cr & \,\,\,\,\, = {{\left( {f – 4h} \right).8} \over {3k.8}} + {{ – \left( {2f – 5h} \right).3} \over {8k.3}} \cr & \,\,\,\,\, – {{8f – 32h – 6f + 15h} \over {24k}} = {{2f – 32k + 15h} \over {24k}} \cr & f)\,\,{{p + 3} \over {2z}} + {{p – 1} \over {6z}} – {{2p + 1} \over {3z}} \cr & \,\,\,\,\, = {{\left( {p + 3} \right).3} \over {2z.3}} + {{p – 1} \over {6z}} – {{\left( {2p + 1} \right).2} \over {3z.2}} \cr & \,\,\,\,\, = {{3p + 9 + p – 1 – 4p – 2} \over {6z}} = {6 \over {6z}} = {1 \over z} \cr} \)