Rút gọn phân thức sau:
a) \({{{x^4} – 4{x^2}} \over {x{{(x + 2)}^2}}}\) ;
b) \({{{x^2} + 2x} \over {{x^2} + 4x + 4}}\).
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\(\eqalign{ & a)\,\,{{{x^4} – 4{x^2}} \over {x{{\left( {x + 2} \right)}^2}}} = {{{x^2}\left( {{x^2} – 4} \right)} \over {x{{\left( {x + 2} \right)}^2}}} = {{{x^2}\left( {x – 2} \right)\left( {x + 2} \right)} \over {x{{\left( {x + 2} \right)}^2}}} \cr & \,\,\,\,\,\, = {{\left[ {x\left( {x + 2} \right)} \right]\left[ {x\left( {x – 2} \right)} \right]} \over {\left[ {x\left( {x + 2} \right)} \right]\left( {x + 2} \right)}} = {{x\left( {x – 2} \right)} \over {x + 2}} \cr & b)\,\,{{{x^2} + 2x} \over {{x^2} + 4x + 4}} = {{x\left( {x + 2} \right)} \over {{{\left( {x + 2} \right)}^2}}} = {{x\left( {x + 2} \right)} \over {\left( {x + 2} \right)\left( {x + 2} \right)}} = {x \over {x + 2}} \cr} \)