\(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=\left(x+4\right)^2\)
ĐKXĐ : \(x\ne0\)
Ta có \(pt\Leftrightarrow8\left(x^2+\frac{1}{x^2}+2\right)+4\left(x^2+\frac{1}{x}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x^2+\frac{1}{x^2}+2\right)=\left(x+4\right)^2\)
Đặt \(x^2+\frac{1}{x^2}=a\) thay vào pt trên ta có :
\(pt\Leftrightarrow8\left(a+2\right)+4a^2-4.a.\left(a+2\right)=\left(x+4\right)^2\)
\(\Leftrightarrow8a+16+4a^2-4a^2-8a=\left(x+4\right)^2\)
\(\Leftrightarrow\left(x+4\right)^2=16\Leftrightarrow\orbr{\begin{cases}x+4=4\\x+4=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=0\left(KTMĐKXĐ\right)\\x=-8\left(TMĐKXĐ\right)\end{cases}}}\)
Vậy \(x=-8\)
\(\)
ko biet vua chia tay nen ko tra loi dc huhu em oi
