Question

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

Answer 1

Mạn phép mk sửa lại đề bài

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

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

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

⇔ ( x + 4)² = 16

⇔ x² + 8x = 0

⇔ x( x + 8) = 0

⇔ x = 0 ( KTM ) hoặc x = -8 ( TM )

KL....