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a) \({{d – 1} \over {3d – 7}} – {1 \over {14 – 6d}}\) ;
b) \({{2n – 5} \over {9n – 6}} – {{m + 3} \over {4 – 6n}}\) ;
c) \({{u + 1} \over {2u – 8}} – {{u + 2} \over {12 – 3u}}\) ;
d) \({{5{a^2}} \over {6a – 6b}} – {{2{a^2}} \over {3b – 3a}}\) ;
e) \({{3x} \over {4y – 2z}} – {{2x} \over {z – 2y}} + {5 \over {3z – 6y}}\) .
Advertisements (Quảng cáo)
\(\eqalign{ & a)\,\,{{d – 1} \over {3d – 7}} – {1 \over {14 – 6d}} = {{d – 1} \over {3d – 7}} + {{ – 1} \over {14 – 6d}} \cr & \,\,\,\,\, = {{\left( {d – 1} \right)\left( {14 – 6d} \right)} \over {\left( {3d – 7} \right)\left( {14 – 6d} \right)}} + {{ – 1\left( {3d – 7} \right)} \over {\left( {14 – 6d} \right)\left( {3d – 7} \right)}} \cr & \,\,\,\, = {{14d – 6{d^2} – 14 + 6d – 3d + 7} \over {\left( {3d – 7} \right)\left( {14 – 6d} \right)}} \cr & \,\,\,\,\, = {{ – 6{d^2} + 17d – 7} \over {\left( {3d – 7} \right)\left( {14 – 6d} \right)}} = {{ – 6{d^2} + 3d + 14d – 7} \over {\left( {3d – 7} \right)\left( {14 – 6d} \right)}} \cr & \,\,\,\,\, = {{ – 3d\left( {2d – 1} \right) + 7\left( {2d – 1} \right)} \over {\left( {3d – 7} \right)\left( {14 – 6d} \right)}} \cr & \,\,\,\,\, = {{\left( {2d – 1} \right)\left( { – 3d + 7} \right)} \over { – \left( { – 3d + 7} \right)\left( {14 – 6d} \right)}} = {{2d – 1} \over { – 14 + 6d}} \cr & b)\,\,{{2n – 5} \over {9n – 6}} – {{m + 3} \over {4 – 6n}} = {{2n – 5} \over {3\left( {3n – 2} \right)}} + {{m + 3} \over {2\left( {3n – 2} \right)}} \cr & \,\,\,\,\, = {{\left( {2n – 5} \right).2} \over {3\left( {3n – 2} \right).2}} + {{\left( {m + 3} \right).3} \over {2\left( {3n – 2} \right).3}} \cr & \,\,\,\, = {{4n – 10 + 3m + 9} \over {6\left( {3n – 2} \right)}} = {{4n + 3m – 1} \over {6\left( {3n – 2} \right)}} \cr & c)\,\,{{u + 1} \over {2u – 8}} – {{u + 2} \over {12 – 3u}} = {{u + 1} \over {2\left( {u – 4} \right)}} + {{u + 2} \over { – \left( {12 – 3u} \right)}} \cr & \,\,\,\,\, = {{u + 1} \over {2\left( {u – 4} \right)}} + {{u + 2} \over {3\left( {u – 4} \right)}} = {{\left( {u + 1} \right).3} \over {2\left( {u – 4} \right).3}} + {{\left( {u + 2} \right).2} \over {3\left( {u – 4} \right).2}} \cr & \,\,\,\,\, = {{3u + 3 + 2u + 4} \over {6\left( {u – 4} \right)}} = {{5u + 7} \over {6\left( {u – 4} \right)}} \cr & d)\,\,{{5{a^2}} \over {6a – 6b}} – {{2{a^2}} \over {3b – 3a}} = {{5{a^2}} \over {6\left( {a – b} \right)}} + {{2{a^2}} \over { – \left( {3b – 3a} \right)}} \cr & \,\,\,\,\, = {{5{a^2}} \over {6\left( {a – b} \right)}} + {{2{a^2}} \over {3\left( {a – b} \right)}} = {{5{a^2}} \over {6\left( {a – b} \right)}} + {{2{a^2}.2} \over {3\left( {a – b} \right).2}} \cr & \,\,\,\,\, = {{5{a^2} + 4{a^2}} \over {6\left( {a – b} \right)}} = {{9{a^2}} \over {6\left( {a – b} \right)}} = {{3{a^2}} \over {2\left( {a – b} \right)}} \cr} \)
\(\eqalign{ & e)\,\,{{3x} \over {4y – 2z}} – {{2x} \over {z – 2y}} + {5 \over {3z – 6y}} \cr & \,\,\,\,\, = {{3x} \over {2\left( {2y – z} \right)}} + {{2x} \over {2y – z}} + {{ – 5} \over {3\left( {2y – z} \right)}} \cr & \,\,\,\,\, = {{3x.3} \over {2\left( {2y – z} \right).3}} + {{2x.6} \over {\left( {2y – z} \right).6}} + {{ – 5.2} \over {3\left( {2y – z} \right).2}} \cr & \,\,\,\,\, = {{9x + 12x – 10} \over {6\left( {2y – z} \right)}} = {{21x – 10} \over {6\left( {2y – z} \right)}} \cr} \)