Question

giải phương trình

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

Answer 1

\(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x+4\right)^2\) ⇔ \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)\left(x^2+\dfrac{1}{x^2}-x^2-\dfrac{1}{x^2}-2\right)=\left(x+4\right)^2\) ⇔ \(8\left(x+\dfrac{1}{x}\right)^2-8\left(x^2+\dfrac{1}{x^2}\right)=\left(x+4\right)^2\) ( x # 0 )

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

⇔ \(x^2+8x=0\)

⇔ \(x=0\left(KTM\right)orx=-8\left(TM\right)\)

KL...............