Question

\(\dfrac{x-x^2}{1-x^2}+\dfrac{1+x}{1+2x+x^2}+\dfrac{1-2x}{x-1}-\dfrac{x^3-1}{1-x^3}\) tại x=5

Answer 1

\(\dfrac{x\left(1-x\right)}{\left(1-x\right)\left(1+x\right)}\)

Answer 2

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

=\(\dfrac{x}{1+x}+\dfrac{1}{1+x}+\dfrac{1-2x}{x-1}+1\)

=\(1+1+\dfrac{1-2x}{x-1}\)

Thay x=5 ta có \(1+1+\dfrac{1-10}{10-1}\)

=2-1=1