Đáp án:

$S = \left\{ { \pm \sqrt 2 } \right\}$

Giải thích các bước giải:

 Ta có:

$\begin{array}{l}
\dfrac{{4{x^2} + 16}}{{{x^2} + 6}} - \dfrac{3}{{{x^2} + 1}} = \dfrac{5}{{{x^2} + 3}} + \dfrac{7}{{{x^2} + 5}}\\
 \Leftrightarrow \dfrac{{4\left( {{x^2} + 6} \right) - 8}}{{{x^2} + 6}} - \dfrac{3}{{{x^2} + 1}} = \dfrac{5}{{{x^2} + 3}} + \dfrac{7}{{{x^2} + 5}}\\
 \Leftrightarrow 4 - \dfrac{8}{{{x^2} + 6}} - \dfrac{3}{{{x^2} + 1}} = \dfrac{5}{{{x^2} + 3}} + \dfrac{7}{{{x^2} + 5}}\\
 \Leftrightarrow \dfrac{8}{{{x^2} + 6}} + \dfrac{3}{{{x^2} + 1}} + \dfrac{5}{{{x^2} + 3}} + \dfrac{7}{{{x^2} + 5}} - 4 = 0\\
 \Leftrightarrow \left( {\dfrac{8}{{{x^2} + 6}} - 1} \right) + \left( {\dfrac{3}{{{x^2} + 1}} - 1} \right) + \left( {\dfrac{5}{{{x^2} + 3}} - 1} \right) + \left( {\dfrac{7}{{{x^2} + 5}} - 1} \right) = 0\\
 \Leftrightarrow \dfrac{{2 - {x^2}}}{{{x^2} + 6}} + \dfrac{{2 - {x^2}}}{{{x^2} + 1}} + \dfrac{{2 - {x^2}}}{{{x^2} + 3}} + \dfrac{{2 - {x^2}}}{{{x^2} + 5}} = 0\\
 \Leftrightarrow \left( {2 - {x^2}} \right)\left( {\dfrac{1}{{{x^2} + 6}} + \dfrac{1}{{{x^2} + 1}} + \dfrac{1}{{{x^2} + 3}} + \dfrac{1}{{{x^2} + 5}}} \right) = 0\\
 \Leftrightarrow 2 - {x^2} = 0\left( {do:\dfrac{1}{{{x^2} + 6}} + \dfrac{1}{{{x^2} + 1}} + \dfrac{1}{{{x^2} + 3}} + \dfrac{1}{{{x^2} + 5}} > 0,\forall x} \right)\\
 \Leftrightarrow {x^2} = 2\\
 \Leftrightarrow x =  \pm \sqrt 2 
\end{array}$

Vậy tập nghiệm của phương trình là: $S = \left\{ { \pm \sqrt 2 } \right\}$

`(4x^2+16)/(x^2+6) - 3/(x^2+1) = 5/(x^2+3) + 7/(x^2+5)`

`⇔ ((4x^2+16)(x^2+1)(x^2+3)(x^2+5)-3(x^2+6)(x^2+3)(x^2+5))/((x^2+6)(x^2+1)(x^2+3)(x^2+5))=(5(x^2+6)(x^2+1)(x^2+5)+7(x^2+6)(x^2+1)(x^2+3))/((x^2+6)(x^2+1)(x^2+3)(x^2+5))`

`⇒ (4x^2+16)(x^2+1)(x^2 + 5) - 3(x^2+6)(x^2+3)(x^2+5) = 5(x^2+6)(x^2+1)(x^2+5)+7(x^2+6)(x^2+1)(x^2+3)`

`⇔ 4x^8 + 49x^6 + 194x^4 + 239x^2 - 30 = 12x^6 + 130x^4 + 394x^2 + 276`

`⇔ 4x^8 + 37x^6 + 64x^4 - 155x^2 - 306 = 0`

Đặt `x^2 = t(t \ge 0)`

`⇔ 4t^4 + 37t^3 + 64t^2 - 155t - 306 = 0`

`⇔ (t-2)(4t^3 + 45t^2 + 154t + 153) = 0`

`⇔`\(\left[ \begin{array}{l}t-2=0\\4t^3+45t^2+154t+153=0\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}t=2(TM)\\t ∈ \mathbb{R}\end{array} \right.\) 

`⇔ x^2 = 2`

`⇔ x = \pm\sqrt{2}`

Vậy `S = {\pm\sqrt{2}}`