ĐKXĐ : \(\forall x\)
Ta có : \(\dfrac{x^2}{x^2+2x+2}+\dfrac{x^2}{x^2-2x+2}-\dfrac{4\left(x^2-5\right)}{x^4+4}=\dfrac{322}{65}\)
\(\Leftrightarrow\dfrac{x^2\left(x^2-2x+2\right)+x^2\left(x^2+2x+2\right)-4\left(x^2-5\right)}{x^4+4}=\dfrac{322}{65}\)
\(\Leftrightarrow\dfrac{x^4-2x^3+2x^2+x^4+2x^3+2x^2-4x^2+20}{x^4+4}=\dfrac{322}{65}\)
\(\Leftrightarrow\dfrac{2x^4+20}{x^4+4}=\dfrac{322}{65}\)
\(\Leftrightarrow65\left(2x^4+20\right)=322\left(x^4+4\right)\)
\(\Leftrightarrow130x^4+1300=322x^4+1288\)
\(\Leftrightarrow192x^4-12=0\)
\(\Leftrightarrow x^4=\dfrac{12}{192}\)
\(\Leftrightarrow x^4=\dfrac{1}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy ...
\(\dfrac{x^2}{x^2+2x+2}+\dfrac{x^2}{x^2-2x+2}-\dfrac{4\left(x^2-5\right)}{x^4+4}=\dfrac{322}{65}\)
\(\Leftrightarrow x^2\left(\dfrac{1}{x^2+2+2x}+\dfrac{1}{x^2+2-2x}\right)-\dfrac{4\left(x^2-5\right)}{x^4+4}=\dfrac{322}{65}\)
\(\Leftrightarrow x^2\left(\dfrac{2x^2+4}{x^4+4}\right)-\dfrac{4\left(x^2-5\right)}{x^4+4}=\dfrac{322}{65}\)
\(\Leftrightarrow\dfrac{2x^4+4x^2-4x^2+20}{x^4+4}=\dfrac{322}{65}\)
\(\Leftrightarrow\dfrac{2x^4+20}{x^4+4}=\dfrac{322}{65}\)
\(\Leftrightarrow65x^4+650=161x^4+644\)
\(\Leftrightarrow96x^4=6\)
\(\Leftrightarrow x^4=\dfrac{1}{16}\)
\(\Rightarrow x=\pm\dfrac{1}{2}\)
