Giải phương trình sau: `8(x+1/x)^2 + 4(x^2+1/x^2)^2 - 4(x^2+1/x^2)(x+1/x)^2 = (x+4)^2`

$ĐKXĐ:x\neq0$

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

$⇔8(x+\dfrac{1}{x})^2+4(x^2+\dfrac{1}{x^2})(x^2+\dfrac{1}{x^2}-x^2-2-\dfrac{1}{x^2})=(x+4)^2$ ( Nhân tử chung $4(x^2+\dfrac{1}{x^2}$)

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

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

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

$⇔16=(x+4)^2$ 

$⇔$ $\left[ \begin{array}{l}x+4=4\\x+4=-4\end{array} \right.$ 

$⇔$ $\left[ \begin{array}{l}x=0\\x=-8\end{array} \right.$ 

Mà: $x\neq0$

$⇔x=-8$

Vậy `S={-8}`