Đáp án:
\(\eqalign{
& a)\,\,x \ne - 4;\,\,x \ne \pm 3;\,\,x \ne - 1 \cr
& b)\,\,A = {{x{{\left( {x + 1} \right)}^2}} \over {\left( {x + 4} \right)\left( {9 - {x^2}} \right)}} \cr} \)
Giải thích các bước giải:
\(\eqalign{
& A = \left( {1 - {4 \over {x + 4}}} \right):{{9 - {x^2}} \over {{x^2} + 2x + 1}} \cr
& a)\,\,x \ne - 4;\,\,x \ne \pm 3;\,\,x \ne - 1 \cr
& b)\,\,A = \left( {1 - {4 \over {x + 4}}} \right):{{9 - {x^2}} \over {{x^2} + 2x + 1}} \cr
& \,\,\,\,\,A = {{x + 4 - 4} \over {x + 4}}.{{{{\left( {x + 1} \right)}^2}} \over {9 - {x^2}}} \cr
& \,\,\,\,A = {{x{{\left( {x + 1} \right)}^2}} \over {\left( {x + 4} \right)\left( {9 - {x^2}} \right)}} \cr} \)
