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Thực hiện tính chia:
a) \(({m^4} – 2{m^2}{n^2} + {n^4}):({m^2} – {n^2})\) ;
b) \((25{x^2} – 81{y^2}):(5x + 9y)\) ;
c) \((8{x^3} + 1):(4{x^2} – 2x + 1)\) .
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\(\eqalign{ & a)\,\,\left( {{m^4} – 2{m^2}{n^2} + {n^4}} \right):\left( {{m^2} – {n^2}} \right) \cr & \,\,\,\,\,\, = \left[ {{{\left( {{m^2}} \right)}^2} – 2{m^2}{n^2} + {{\left( {{n^2}} \right)}^2}} \right]:\left( {{m^2} – {n^2}} \right) \cr & \,\,\,\,\,\, = {\left( {{m^2} – {n^2}} \right)^2}:\left( {{m^2} – {n^2}} \right) \cr & \,\,\,\,\,\, = {{{{\left( {{m^2} – {n^2}} \right)}^2}} \over {{m^2} – {n^2}}} = {m^2} – {n^2} \cr & b)\,\,\left( {25{x^2} – 81{y^2}} \right):\left( {5x + 9y} \right) \cr & \,\,\,\,\,\, = {{25{x^2} – 81{y^2}} \over {5x + 9y}} = {{{{\left( {5x} \right)}^2} – {{\left( {9y} \right)}^2}} \over {5x + 9y}} \cr & \,\,\,\,\,\, = {{\left( {5x – 9y} \right)\left( {5x + 9y} \right)} \over {5x + 9y}} = 5x – 9y \cr & c)\,\,\left( {8{x^3} + 1} \right):\left( {4{x^2} – 2x + 1} \right) \cr & \,\,\,\,\,\, = {{8{x^3} + 1} \over {4{x^2} – 2x + 1}} = {{{{\left( {2x} \right)}^3} + {1^3}} \over {4{x^2} – 2x + 1}} \cr & \,\,\,\,\,\, = {{\left( {2x + 1} \right)\left( {4{x^2} – 2x + 1} \right)} \over {4{x^2} – 2x + 1}} = 2x + 1 \cr} \)