Giải thích các bước giải:

a.ĐKXĐ: $x\ne\pm1$ 

$\dfrac1{x+1}-\dfrac{2x}{x-1}+\dfrac{x+3}{x^2-1}$

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

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

$=\dfrac{-2x^2+2}{x^2-1}$

$=-2$

b.ĐKXĐ: $x\ne\pm\dfrac12$

$\dfrac2{2x+1}-\dfrac1{2x-1}+\dfrac2{4x^2-1}$

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

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

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

$=\dfrac{1}{2x+1}$

c.ĐKXĐ: $x\ne 0, x\ne\pm\dfrac32$

$\dfrac7{8x^2-18}+\dfrac1{2x^2+3x}-\dfrac1{4x-6}$

$=\dfrac{7x}{2x\left(2x+3\right)\left(2x-3\right)}+\dfrac{2\left(2x-3\right)}{2x\left(2x+3\right)\left(2x-3\right)}-\dfrac{x\left(2x+3\right)}{2x\left(2x+3\right)\left(2x-3\right)}$

$=\dfrac{7x+2\left(2x-3\right)-x\left(2x+3\right)}{2x\left(2x+3\right)\left(2x-3\right)}$

$=\dfrac{-2x^2+8x-6}{2x\left(2x+3\right)\left(2x-3\right)}$

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

d.ĐKXĐ: $x\ne -1$

$\dfrac{3x^2+5x+14}{x^3+1}+\dfrac{x-1}{x^2-x+1}-\dfrac4{x+1}$

$=\dfrac{3x^2+5x+14}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{4\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}$

$=\dfrac{3x^2+5x+14+\left(x-1\right)\left(x+1\right)-4\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}$

$=\dfrac{9x+9}{\left(x+1\right)\left(x^2-x+1\right)}$

$=\dfrac{9}{x^2-x+1}$