Đáp án:

 a) $\dfrac{x-3}{x-5}$ + $\dfrac{1}{x}$ = $\dfrac{x+5}{x(x-5)}$

ĐKXĐ : x $\neq$ 0 ; x $\neq$ 5

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

⇒ x(x-3) + x-5 = x+5

⇔ x² -3x + x - 5 =x + 5

⇔ x² -3x +x -x -5 -5 = 0

⇔ x² -3x - 10 =0

⇔ x² -5x + 2x -10 =0

⇔ -x(-x+5) -2(x+5)=0

⇔ (-x+5)(-x-2)=0

⇔\(\left[ \begin{array}{l}-x+5=0\\-x-2=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=5 (KTM)\\x=-2(TM)\end{array} \right.\) 

Vậy phương trình có tập nghiệm S={ -2}

b) $\dfrac{1+x}{1-x}$ + 3 = $\dfrac{3-x}{1-x}$

ĐKXĐ : x $\neq$ 1

⇔ $\dfrac{1+x}{1-x}$ + $\dfrac{3(1-x)}{1-x}$ = $\dfrac{3-x}{1-x}$

⇒ 1+ x +3(1-x) = 3-x

⇔ 1 + x + 3 -3x = 3-x

⇔ x -3x +x = 3 -3 -1

⇔ -x = -1

⇔ x = 1 (KTM)

⇔ x vô nghiệm

c) $\dfrac{(2x+1)²}{x²-1}$ + $\dfrac{x-1}{x+1}$ = $\dfrac{2x+5 + 3}{1 -x }$

ĐKXĐ : x $\neq$ ±1

⇔ $\dfrac{(2x+1)²}{(x-1)(x+1)}$ + $\dfrac{(x-1)²}{x+1}$ = $\dfrac{-(2x+8)(x+1)}{(x-1)(x+1)}$

⇒(2x+1)² + (x-1)² = -(2x+8)(x+1)

⇔ 4x² + 4x +1 + x² -2x +1 = -2x² -2x -8x -8

⇔ 4x² +x +2x² +4x -2x +2x +8x +1 +1 +8=0

⇔ 7x² +12x +10 =0

⇔ (√7x)²+2 .  $\dfrac{6√7}{7}$ . √7x + $\dfrac{36}{7}$+ $\dfrac{34}{7}$ =0

⇔ (√7x + $\dfrac{6√7}{7}$)² + $\dfrac{34}{7}$ = 0

⇔ (√7x + $\dfrac{6√7}{7}$)² ≥ 0 ∀x

⇔(√7x + $\dfrac{6√7}{7}$)² + $\dfrac{34}{7}$ >0 ∀x

⇒x vô nghiệm