Question

Giải phương trình:

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

Answer 1

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

<=>\(8\left(x+\dfrac{1}{x}\right)^2+4\left[\left(x+\dfrac{1}{x}\right)^2-2\right]^2-4\left[\left(x+\dfrac{1}{x}\right)^2-2\right]\left(x+\dfrac{1}{x}\right)^2=\left(x+4\right)^2\) Đặt \(\left(x+\dfrac{1}{x}\right)^2\) = a => (*) trở thành: \(8a+4\left(a-2\right)^2-4a\left(a-2\right)=x^2+8x+16\) <=> \(8a+4a^2-16a+16-4a^2-8a=x^2+8x+16\) <=> \(x^2+8x+16=16\) <=> \(x^2+8x=0\) <=> \(x\left(x+8\right)=0\) <=> \(\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\) Vậy ..................................................