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Trong mỗi trường hợp sau hãy tìm phân thức Q thỏa mãn điều kiện :
a. \({1 \over {{x^2} + x + 1}} – Q = {1 \over {x – {x^2}}} + {{{x^2} + 2x} \over {{x^3} – 1}}\)
b. \({{2x – 6} \over {{x^3} – 3{x^2} – x + 3}} + Q = {6 \over {x – 3}} – {{2{x^2}} \over {1 – {x^2}}}\)
a. \({1 \over {{x^2} + x + 1}} – Q = {1 \over {x – {x^2}}} + {{{x^2} + 2x} \over {{x^3} – 1}}\)
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\(\eqalign{ & Q = {1 \over {{x^2} + x + 1}} – {1 \over {x – {x^2}}} – {{{x^2} + 2x} \over {{x^3} – 1}} \cr & Q = {1 \over {{x^2} + x + 1}} + {1 \over {x\left( {x – 1} \right)}} – {{{x^2} + 2x} \over {\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}} \cr & Q = {{x\left( {x – 1} \right) + {x^2} + x + 1 – x\left( {{x^2} + 2x} \right)} \over {x\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}} \cr & Q = {{{x^2} – x + {x^2} + x + 1 – {x^3} – 2{x^2}} \over {x\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}} = {{1 – {x^3}} \over {x\left( {{x^3} – 1} \right)}} = {{ – \left( {{x^3} – 1} \right)} \over {x\left( {{x^3} – 1} \right)}} \cr & Q = – {1 \over x} \cr} \)
b. \({{2x – 6} \over {{x^3} – 3{x^2} – x + 3}} + Q = {6 \over {x – 3}} – {{2{x^2}} \over {1 – {x^2}}}\)
\(\eqalign{ & Q = {6 \over {x – 3}} + {{2{x^2}} \over {{x^2} – 1}} – {{2x – 6} \over {{x^3} – 3{x^2} – x + 3}} \cr & Q = {6 \over {x – 3}} + {{2{x^2}} \over {{x^2} – 1}} – {{2x – 6} \over {\left( {x – 3} \right)\left( {{x^2} – 1} \right)}} \cr & Q = {{6\left( {{x^2} – 1} \right) + 2{x^2}\left( {x – 3} \right) – \left( {2x – 6} \right)} \over {\left( {x – 3} \right)\left( {{x^2} – 1} \right)}} \cr & Q = {{6{x^2} – 6 + 2{x^3} – 6{x^2} – 2x + 6} \over {\left( {x – 3} \right)\left( {{x^2} – 1} \right)}} = {{2{x^3} – 2x} \over {\left( {x – 3} \right)\left( {{x^2} – 1} \right)}} = {{2x\left( {{x^2} – 1} \right)} \over {\left( {x – 3} \right)\left( {{x^2} – 1} \right)}} \cr & Q = {{2x} \over {x – 3}} \cr} \)