Giai pt $\frac{4x^2+16}{x^2+6}-\frac{3}{x^2+1}=\frac{5}{x^2+3}+\frac{7}{x^2+5}$

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