Bài tập 9 trang 79 Tài liệu dạy – học Toán 8 tập 1

Đề bài

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) \({{3x} \over {4y - 2z}} - {{2x} \over {z - 2y}} + {5 \over {3z - 6y}}\) .

Lời giải chi tiết

\(\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} \)

\(\eqalign{  & e)\,\,{{3x} \over {4y - 2z}} - {{2x} \over {z - 2y}} + {5 \over {3z - 6y}}  \cr  & \,\,\,\,\, = {{3x} \over {2\left( {2y - z} \right)}} + {{2x} \over {2y - z}} + {{ - 5} \over {3\left( {2y - z} \right)}}  \cr  & \,\,\,\,\, = {{3x.3} \over {2\left( {2y - z} \right).3}} + {{2x.6} \over {\left( {2y - z} \right).6}} + {{ - 5.2} \over {3\left( {2y - z} \right).2}}  \cr  & \,\,\,\,\, = {{9x + 12x - 10} \over {6\left( {2y - z} \right)}} = {{21x - 10} \over {6\left( {2y - z} \right)}} \cr} \)