Question

Tính

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

Answer 1

ĐKXĐ bạn tự tìm nhé

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

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

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

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

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

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