\(N=\left(\dfrac{3x}{2x-2}-\dfrac{3x}{2x+2}-\dfrac{3x-2}{x^2-1}\right)\cdot\dfrac{4x^2-4}{3}\left(x\ne\pm1\right)\)
\(=\left[\dfrac{3x}{2\left(x-1\right)}-\dfrac{3x}{2\left(x+1\right)}-\dfrac{3x-2}{\left(x-1\right)\left(x+1\right)}\right]\cdot\dfrac{4x^2-4}{3}\)
\(=\left[\dfrac{3x\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\dfrac{3x\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}-\dfrac{2\cdot\left(3x-2\right)}{2\left(x-1\right)\left(x+1\right)}\right]\cdot\dfrac{4\left(x^2-1\right)}{3}\)
\(=\dfrac{3x^2+3x-3x^2+3x-6x+4}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{3}\)
\(=\dfrac{4\cdot2}{3}\)
\(=\dfrac{8}{3}\)
Vậy giá trị của \(N\) không phụ thuộc vào biến \(x\).
\(N=\left(\dfrac{3x}{2x-2}-\dfrac{3x}{2x+2}-\dfrac{3x-2}{x^2-1}\right)\cdot\dfrac{4x^2-4}{3}\\ =\left(\dfrac{3x}{2\left(x-1\right)}-\dfrac{3x}{2\left(x+1\right)}-\dfrac{3x-2}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{4x^2-4}{3}\\ =\dfrac{3x\left(x+1\right)-3x\left(x-1\right)-2\left(3x-2\right)}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{4x^2-4}{3}\\ =\dfrac{\left(3x^2+3x\right)-\left(3x^2-3x\right)-\left(6x-4\right)}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{4x^2-4}{3}\)
\(=\dfrac{3x^2+3x-3x^2+3x-6x+4}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{3}\\ =\dfrac{4}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{3}\\ =\dfrac{4\cdot2}{3}\)
`=8/3`
`->N` không phụ thuộc vào biến `x`
`color{blue}{ @ A_ri}`
