Question

Thực hiện phép tính\(\left(\dfrac{x+1}{2\left(x-1\right)}+\dfrac{3}{x^2-1}-\dfrac{x+3}{2\left(x+1\right)}\right):\dfrac{x}{x^2-1}\)

Answer 1

Lâu lắm r ms qua toán 8!

ĐKXĐ: \(x\ne\pm1;0\)

Ta có: \(\left[\dfrac{x+1}{2\left(x-1\right)}+\dfrac{3}{x^2-1}-\dfrac{x+3}{2\left(x+1\right)}\right]:\dfrac{x}{x^2-1}\)

\(=\left[\dfrac{\left(x+1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}+\dfrac{3.2}{2\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right]\)

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

\(=\dfrac{x^2+2x+1+6-x^2+x-3x+3}{2\left(x-1\right)\left(x+1\right)}:\dfrac{x}{x^2-1}\)

\(=\dfrac{10}{2\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{x}\) \(=\dfrac{5}{x}\)