Rút gọn:
\(P = {{\left( {x + 1} \right)\left( {4{x^2} - 4x + 1} \right) + \left( {x - 1} \right)\left( {4{x^2} - 4x + 1} \right)} \over {\left( {x + 3} \right)\left( {x - 1} \right) - {x^2} - 1}}\) (với \(\left( {2x - 1} \right) \ne 0\) )
Giải chi tiết:
\(\eqalign{& P = {{\left( {x + 1} \right)\left( {4{x^2} - 4x + 1} \right) + \left( {x - 1} \right)\left( {4{x^2} - 4x + 1} \right)} \over {\left( {x + 3} \right)\left( {x - 1} \right) - {x^2} - 1}} \cr & P = {{\left( {x + 1 + x - 1} \right)\left( {{{\left( {2x} \right)}^2} - 2.2x.1 + {1^2}} \right)} \over {{x^2} - x + 3x - 3 - {x^2} - 1}} \cr & P = {{2x{{\left( {2x - 1} \right)}^2}} \over {2x - 4}} = {{2x{{\left( {2x - 1} \right)}^2}} \over {2\left( {x - 2} \right)}} \cr & P = {{x{{\left( {2x - 1} \right)}^2}} \over {x - 2}} \cr} \)