ta có:
A = \(\left(\dfrac{x+3}{2x+2}+\dfrac{3}{1-x^2}-\dfrac{x+1}{2x-2}\right):\dfrac{3}{2x^2-2}\)
= \(\left(\dfrac{x+3}{2\left(x+1\right)}-\dfrac{3}{x^2-1}-\dfrac{x+1}{2\left(x-1\right)}\right):\dfrac{3}{2\left(x^2-1\right)}\)
= \(\left(\dfrac{x+3}{2\left(x+1\right)}-\dfrac{3}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+1}{2\left(x-1\right)}\right):\dfrac{3}{2\left(x-1\right)\left(x+1\right)}\)
= \(\left(\dfrac{\left(x+3\right)\left(x-1\right)}{2\left(x+1\right)\left(x-1\right)}-\dfrac{6}{2\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}\right):\dfrac{3}{2\left(x-1\right)\left(x+1\right)}\)
= \(\left(\dfrac{x^2-x+3x-3-6-x^2-2x-1}{2\left(x+1\right)\left(x-1\right)}\right):\dfrac{3}{2\left(x-1\right)\left(x+1\right)}\)
= \(-\dfrac{10}{2\left(x+1\right)\left(x-1\right)}.\dfrac{2\left(x+1\right)\left(x-1\right)}{3}\)
= \(-\dfrac{10}{3}\)
Vậy phương trình trên ko phụ thuộc vào biến
