Question

Thực hiện phép tính:

\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)

Answer 1

=\(\dfrac{8}{\left(x^2+3\right)\left(x+1\right)\left(x-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)

=\(\dfrac{8}{\left(x^2+3\right)\left(x+1\right)\left(x-1\right)}+\dfrac{2\left(x+1\right)\left(x-1\right)}{\left(x^2+3\right)\left(x+1\right)\left(x-1\right)}+\dfrac{\left(x^2+3\right)\left(x-1\right)}{\left(x^2+3\right)\left(x+1\right)\left(x-1\right)}\)

=\(\dfrac{8+2\left(x+1\right)\left(x-1\right)+\left(x^2+3\right)\left(x-1\right)}{\left(x^2+3\right)\left(x+1\right)\left(x-1\right)}\)

=\(8+2\left(x-1\right)\)

=\(8+2x-2\)

=2x+6