Question

thực hiện phép tính

 

Answer 1

\(\dfrac{1}{1+a}\) +\(\dfrac{1}{a-1}+\dfrac{3a}{1-a^2}\)

=\(\dfrac{1}{1+a}-\dfrac{1}{1-a}+\dfrac{3a}{1-a^2}\) 

=\(\dfrac{1}{1+a}-\dfrac{1}{1-a}+\dfrac{3a}{\left(1+a\right).\left(1-a\right)}\)  

MTC:(1+a).(1-a)

=\(\dfrac{1.\left(1-a\right)}{\left(1+a\right).\left(1-a\right)}-\dfrac{1.\left(1+a\right)}{\left(1+a\right).\left(1-a\right)}+\dfrac{3a}{\left(1+a\right).\left(1-a\right)}\) 

=\(\dfrac{1-a-1-a+3a}{\left(1+a\right).\left(1-a\right)}\)

=\(\dfrac{3a}{\left(1+a\right).\left(1-a\right)}\)