rút gọn phân thức sau A= x + 1 / 3x - x^2 : ( 3+ x/3 -x - 3x/ 3+x - 12x^2/x^2 - 9 )

Đáp án:

$\texttt{d.a.r.i}$

`A= (x + 1)/(3x - x^2) : ( [3+ x]/(3 -x )-(3x)/(3+x)- (12x^2)/(x^2 - 9))` 

`A= (x + 1)/(x.(3 - x)): ( [3+ x]/(3 -x )-(3x)/(3+x) +  (12x^2)/[(3 - x).(3 + x)))` 

`A= (x + 1)/(x.(3 - x)): ( [(3 + x).(3 +x)]/[(3 - x).(3 + x)]-(3x.(3 - x))/[(3 - x).(3 + x)] +  (12x^2)/[(3 - x).(3 + x)])` 

`A= (x + 1)/(x.(3 - x)): ( [x^2 + 6x + 9]/[(3 - x).(3 + x)]-(-3x^2 + 9x)/[(3 - x).(3 + x)] +  (12x^2)/[(3 - x).(3 + x)])` 

`A= (x + 1)/(x.(3 - x)): [x^2 + 6x + 9 + 3x^2 - 9x + 12x^2]/[(3 - x).(3 + x)]` 

`A= (x + 1)/(x.(3 - x)): [16x^2 - 3x + 9]/[(3 - x).(3 + x)]` 

`A= (x + 1)/(x.(3 - x)) . [(3 - x).(3 + x)]/[16x^2 - 3x + 9]`

`A = [(x + 1).(x + 3)]/[x.(16x^2 - 3x + 9)]`