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) \({{y + 1} \over {y – 1}} + {{y + 2} \over {y – 2}} + {y \over {{y^2} – 3y + 2}}\) ;
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e) \({{{1 \over {b + 2}} + {1 \over {b – 5}}} \over {{{2{b^2} – b – 3} \over {{b^2} – 3b – 10}}}}\)
f) \(\left( {{{2s} \over {2s + 1}} – 1} \right):\left( {1 + {{2s} \over {1 – 2s}}} \right)\) .
\(\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} \)
\(\eqalign{ & e)\,\,{{{1 \over {b + 2}} + {1 \over {b – 5}}} \over {{{2{b^2} – b – 3} \over {{b^2} – 3b – 10}}}} = \left( {{1 \over {b + 2}} + {1 \over {b – 5}}} \right):{{2{b^2} – b – 3} \over {{b^2} – 3b – 10}} \cr & \,\,\,\,\, = {{b – 5 + b + 2} \over {\left( {b + 2} \right)\left( {b – 5} \right)}}.{{{b^2} – 3b – 10} \over {2{b^2} – b – 3}} = {{2b – 3} \over {\left( {b + 2} \right)\left( {b – 5} \right)}}.{{{b^2} – 5b + 2b – 10} \over {2{b^2} + 2b – 3b – 3}} \cr & \,\,\,\,\, = {{2b – 3} \over {\left( {b + 2} \right)\left( {b – 5} \right)}}.{{\left( {b – 5} \right)\left( {b + 2} \right)} \over {\left( {b + 1} \right)\left( {2b – 3} \right)}} = {{\left( {2b – 3} \right)\left( {b – 5} \right)\left( {b + 2} \right)} \over {\left( {b + 2} \right)\left( {b – 5} \right)\left( {b + 1} \right)\left( {2b – 3} \right)}} \cr & \,\,\,\,\, = {1 \over {b + 1}} \cr & f)\,\,\left( {{{2s} \over {2s + 1}} – 1} \right):\left( {1 + {{2s} \over {1 – 2s}}} \right) = {{2s – \left( {2s + 1} \right)} \over {2s + 1}}:{{1 – 2s + 2s} \over {1 – 2s}} \cr & \,\,\,\,\,\, = {{ – 1} \over {2s + 1}}.{{1 – 2s} \over 1} = {{2s – 1} \over {2s + 1}} \cr} \)