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. \({{2x - 6} \over {{x^3} - 3{x^2} - x + 3}} + Q = {6 \over {x - 3}} - {{2{x^2}} \over {1 - {x^2}}}\)
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a. \({1 \over {{x^2} + x + 1}} - Q = {1 \over {x - {x^2}}} + {{{x^2} + 2x} \over {{x^3} - 1}}\)
\(\eqalign{ & Q = {1 \over {{x^2} + x + 1}} - {1 \over {x - {x^2}}} - {{{x^2} + 2x} \over {{x^3} - 1}} \cr & Q = {1 \over {{x^2} + x + 1}} + {1 \over {x\left( {x - 1} \right)}} - {{{x^2} + 2x} \over {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} \cr & Q = {{x\left( {x - 1} \right) + {x^2} + x + 1 - x\left( {{x^2} + 2x} \right)} \over {x\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} \cr & Q = {{{x^2} - x + {x^2} + x + 1 - {x^3} - 2{x^2}} \over {x\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = {{1 - {x^3}} \over {x\left( {{x^3} - 1} \right)}} = {{ - \left( {{x^3} - 1} \right)} \over {x\left( {{x^3} - 1} \right)}} \cr & Q = - {1 \over x} \cr} \)
b. \({{2x - 6} \over {{x^3} - 3{x^2} - x + 3}} + Q = {6 \over {x - 3}} - {{2{x^2}} \over {1 - {x^2}}}\)
\(\eqalign{ & Q = {6 \over {x - 3}} + {{2{x^2}} \over {{x^2} - 1}} - {{2x - 6} \over {{x^3} - 3{x^2} - x + 3}} \cr & Q = {6 \over {x - 3}} + {{2{x^2}} \over {{x^2} - 1}} - {{2x - 6} \over {\left( {x - 3} \right)\left( {{x^2} - 1} \right)}} \cr & Q = {{6\left( {{x^2} - 1} \right) + 2{x^2}\left( {x - 3} \right) - \left( {2x - 6} \right)} \over {\left( {x - 3} \right)\left( {{x^2} - 1} \right)}} \cr & Q = {{6{x^2} - 6 + 2{x^3} - 6{x^2} - 2x + 6} \over {\left( {x - 3} \right)\left( {{x^2} - 1} \right)}} = {{2{x^3} - 2x} \over {\left( {x - 3} \right)\left( {{x^2} - 1} \right)}} = {{2x\left( {{x^2} - 1} \right)} \over {\left( {x - 3} \right)\left( {{x^2} - 1} \right)}} \cr & Q = {{2x} \over {x - 3}} \cr} \)