Question

Giải PT

8( x + \(\dfrac{1}{x}\) )² + 4( x² + \(\dfrac{1}{x^2}\)) ^{2 _} 4 (x² + \(\dfrac{1}{x^2}\))( x + \(\dfrac{1}{x}\))² = ( x + 4 )²

Answer 1

Ta có: 8\(\left(x+\dfrac{1}{x}\right)^2\)+4\(\left(x^2+\dfrac{1}{x^2}\right)^2\)\(\left(x+\dfrac{1}{x}\right)^2\)=(x+4)²

ĐKXĐ: x khác 0

<=>8\(\left(x+\dfrac{1}{x}\right)^2\)+4\(\left(x^2+\dfrac{1}{x^2}\right)\)\(\left(x^2+\dfrac{1}{x^2}-x^2-2-\dfrac{1}{x^2}\right)\)=(x+4)²

<=>8\(\left(x+\dfrac{1}{x}\right)^2-8\left(x^2+\dfrac{1}{x^2}\right)=\left(x+4\right)^2\)

<=>8\(\left(x^2+2+\dfrac{1}{x^2}-x^2-\dfrac{1}{x^2}\right)\)=(x+4)²

=>(x+4)²=16

Vậy có 2 TH:

+) x+4=4 => x=0(KTMĐKXĐ)

+)x+4=-4 => x=-8(TMĐKXĐ)

Vậy tập nghiệm của phương trình S={-8}