Ans:

$6)$

$\dfrac{x+1}{x-1} - \dfrac{x-1}{x+1} + \dfrac{x^2-4x-1}{x^2-1}$

$= \dfrac{(x+1)^2 - (x-1)^2 + (x^2-4x-1)}{(x-1)(x+1)}$

$= \dfrac{(x^2+2x+1) - (x^2-2x+1) + x^2 - 4x - 1}{(x-1)(x+1)}$

$= \dfrac{4x + x^2 - 4x - 1}{(x-1)(x+1)}$

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

$7)$

$\dfrac{1}{2x-3} - \dfrac{1}{2x+3} - \dfrac{10x+9}{9-4x^2}$

$= \dfrac{1}{2x-3} - \dfrac{1}{2x+3} + \dfrac{10x+9}{4x^2-9}$

$= \dfrac{(2x+3) - (2x-3) + (10x+9)}{(2x-3)(2x+3)}$

$= \dfrac{10x + 15}{(2x-3)(2x+3)}$

$= \dfrac{5(2x+3)}{(2x-3)(2x+3)} = \dfrac{5}{2x-3}$

$8)$

$\dfrac{2x+1}{2x^2-x} + \dfrac{32x^2}{1-4x^2} + \dfrac{1-2x}{2x^2+x}$

$= \dfrac{2x+1}{x(2x-1)} - \dfrac{32x^2}{(2x-1)(2x+1)} + \dfrac{1-2x}{x(2x+1)}$

$= \dfrac{(2x+1)^2 - 32x^2 \cdot x + (1-2x)(2x-1)}{x(2x-1)(2x+1)}$

$= \dfrac{(4x^2+4x+1) - 32x^3 - (2x-1)^2}{x(4x^2-1)}$

$= \dfrac{4x^2+4x+1 - 32x^3 - (4x^2-4x+1)}{x(4x^2-1)}$

$= \dfrac{8x - 32x^3}{x(4x^2-1)}$

$= \dfrac{-8x(4x^2-1)}{x(4x^2-1)} = -8$

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