Đề bài
Thực hiện phép tính và rút gọn:
a] \[{5 \over {{x^2} - 3x - 28}} + {7 \over {2x - 14}}\] ;
b] \[{{d - 4} \over {{d^2} + 2d - 8}} - {{d + 2} \over {{d^2} - 16}}\] ;
c] \[{{{m^2} + n} \over {{m^2} - {n^2}}} + {m \over {n - m}} + {n \over {m + n}}\] ;
d] \[{{y + 1} \over {y - 1}} + {{y + 2} \over {y - 2}} + {y \over {{y^2} - 3y + 2}}\] ;
e] \[{{{1 \over {b + 2}} + {1 \over {b - 5}}} \over {{{2{b^2} - b - 3} \over {{b^2} - 3b - 10}}}}\]
f] \[\left[ {{{2s} \over {2s + 1}} - 1} \right]:\left[ {1 + {{2s} \over {1 - 2s}}} \right]\] .
Lời giải chi tiết
\[\eqalign{ & a]\,\,{5 \over {{x^2} - 3x - 28}} + {7 \over {2x - 14}} \cr & \,\,\,\,\, = {5 \over {\left[ {x - 7} \right]\left[ {x + 4} \right]}} + {7 \over {2\left[ {x - 7} \right]}} \cr & \,\,\,\,\, = {{5.2} \over {2\left[ {x - 7} \right]\left[ {x + 4} \right]}} + {{7\left[ {x + 4} \right]} \over {2.\left[ {x - 7} \right]\left[ {x + 4} \right]}} \cr & \,\,\,\,\, = {{10 + 7x + 28} \over {2\left[ {x - 7} \right]\left[ {x + 4} \right]}} = {{7x + 38} \over {2\left[ {x - 7} \right]\left[ {x + 4} \right]}} \cr & b]\,\,{{d - 4} \over {{d^2} + 2d - 8}} - {{d + 2} \over {{d^2} - 16}} = {{d - 4} \over {\left[ {d + 4} \right]\left[ {d - 2} \right]}} + {{ - \left[ {d + 2} \right]} \over {\left[ {d - 4} \right]\left[ {d + 4} \right]}} \cr & \,\,\,\,\, = {{{{\left[ {d - 4} \right]}^2} - \left[ {d + 2} \right]\left[ {d - 2} \right]} \over {\left[ {d + 4} \right]\left[ {d - 2} \right]\left[ {d - 4} \right]}} = {{{d^2} - 8d + 16 - {d^2} + 4} \over {\left[ {d + 4} \right]\left[ {d - 2} \right]\left[ {d - 4} \right]}} \cr & \,\,\,\,\, = {{ - 8d + 20} \over {\left[ {d + 4} \right]\left[ {d - 2} \right]\left[ {d - 4} \right]}} \cr & c]\,\,{{{m^2} + {n^2}} \over {{m^2} - {n^2}}} + {m \over {n - m}} + {n \over {m + n}} = {{{m^2} + {n^2}} \over {\left[ {m - n} \right]\left[ {m + n} \right]}} + {{ - m} \over {m - n}} + {n \over {m + n}} \cr & \,\,\,\,\, = {{{m^2} + {n^2}} \over {\left[ {m - n} \right]\left[ {m + n} \right]}} + {{ - m\left[ {m + n} \right]} \over {\left[ {m - n} \right]\left[ {m + n} \right]}} + {{n\left[ {m - n} \right]} \over {\left[ {m - n} \right]\left[ {m + n} \right]}} \cr & \,\,\,\,\, = {{{m^2} + {n^2} - {m^2} - mn + mn - {n^2}} \over {\left[ {m - n} \right]\left[ {m + n} \right]}} = {0 \over {\left[ {m - n} \right]\left[ {m + n} \right]}} = 0 \cr & d]\,\,{{y + 1} \over {y - 1}} + {{y + 2} \over {y - 2}} + {y \over {{y^2} - 3y + 2}} = {{y + 1} \over {y - 1}} + {{y + 2} \over {y - 2}} + {y \over {\left[ {y - 2} \right]\left[ {y - 1} \right]}} \cr & \,\,\,\,\, = {{\left[ {y + 1} \right]\left[ {y - 2} \right]} \over {\left[ {y - 2} \right]\left[ {y - 1} \right]}} + {{\left[ {y + 2} \right]\left[ {y - 1} \right]} \over {\left[ {y - 2} \right]\left[ {y - 1} \right]}} + {y \over {\left[ {y - 2} \right]\left[ {y - 1} \right]}} \cr & \,\,\,\,\, = {{{y^2} - 2y + y - 2 + {y^2} - y + 2y - 2 + y} \over {\left[ {y - 2} \right]\left[ {y - 1} \right]}} \cr & \,\,\,\,\, = {{2{y^2} + y - 4} \over {\left[ {y - 2} \right]\left[ {y - 1} \right]}} \cr} \]
\[\eqalign{ & e]\,\,{{{1 \over {b + 2}} + {1 \over {b - 5}}} \over {{{2{b^2} - b - 3} \over {{b^2} - 3b - 10}}}} = \left[ {{1 \over {b + 2}} + {1 \over {b - 5}}} \right]:{{2{b^2} - b - 3} \over {{b^2} - 3b - 10}} \cr & \,\,\,\,\, = {{b - 5 + b + 2} \over {\left[ {b + 2} \right]\left[ {b - 5} \right]}}.{{{b^2} - 3b - 10} \over {2{b^2} - b - 3}} = {{2b - 3} \over {\left[ {b + 2} \right]\left[ {b - 5} \right]}}.{{{b^2} - 5b + 2b - 10} \over {2{b^2} + 2b - 3b - 3}} \cr & \,\,\,\,\, = {{2b - 3} \over {\left[ {b + 2} \right]\left[ {b - 5} \right]}}.{{\left[ {b - 5} \right]\left[ {b + 2} \right]} \over {\left[ {b + 1} \right]\left[ {2b - 3} \right]}} = {{\left[ {2b - 3} \right]\left[ {b - 5} \right]\left[ {b + 2} \right]} \over {\left[ {b + 2} \right]\left[ {b - 5} \right]\left[ {b + 1} \right]\left[ {2b - 3} \right]}} \cr & \,\,\,\,\, = {1 \over {b + 1}} \cr & f]\,\,\left[ {{{2s} \over {2s + 1}} - 1} \right]:\left[ {1 + {{2s} \over {1 - 2s}}} \right] = {{2s - \left[ {2s + 1} \right]} \over {2s + 1}}:{{1 - 2s + 2s} \over {1 - 2s}} \cr & \,\,\,\,\,\, = {{ - 1} \over {2s + 1}}.{{1 - 2s} \over 1} = {{2s - 1} \over {2s + 1}} \cr} \]