Mày nhìn cái chóa j

\(\Leftrightarrow8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)\left[\left(x^2+\frac{1}{x^2}\right)-\left(x+\frac{1}{x}\right)^2\right]=\left(x+4\right)^2.ĐKXĐ:x\ne0\)

\(\Leftrightarrow8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)\left(x^2+\frac{1}{x^2}-x^2-2-\frac{1}{x^2}\right)=\left(x+4\right)^2\)

\(\Leftrightarrow8\left(x+\frac{1}{x}\right)^2-8\left(x^2+\frac{1}{x^2}\right)=\left(x+4\right)^2\)

\(\Leftrightarrow8\left[\left(x+\frac{1}{x}\right)^2-\left(x^2+\frac{1}{x^2}\right)\right]=\left(x+4\right)^2\)

\(\Leftrightarrow8\left(x^2+2+\frac{1}{x^2}-x^2+\frac{1}{x^2}\right)=\left(x+4\right)^2\)

\(\Leftrightarrow16=\left(x+4\right)^2\)

\(\Leftrightarrow x^2+8x+16=16\)

\(\Leftrightarrow x^2+8x=0\)

\(\Leftrightarrow x\left(x+8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\left(l\right)\\x=-8\left(n\right)\end{cases}}\)

V...\(S=\left\{-8\right\}\)

^^

bạn ghi sai đề ở chỗ \(\left(x+\frac{1}{x}\right)^2\)chứ ko phải \(\left(x+\frac{1}{x^2}\right)^2\)nhé

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

\(\Leftrightarrow4\left(x+\frac{1}{x}\right)^2\left(x^2+\frac{1}{x^2}+2\right)=\left(x+4\right)^2\)

\(\Leftrightarrow4\left(x+\frac{1}{x}\right)^2\left(x+\frac{1}{x}\right)^2=\left(x+4\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}2\left(x+\frac{1}{x}\right)^2=x+4\\2\left(x+\frac{1}{x}\right)^2=-x-4\end{cases}}\)

Tới đây thì đơn giản rồi làm tiếp nhé:

Bạn nhân lần lượt ra, sau đó rút gọn, sau một hồi sẽ được:

     \(\frac{4\left(x^2+1\right)^4}{x^4}=\left(x+4\right)^2\)

\(\Leftrightarrow\frac{4\left(x^2+1\right)^2}{x^2}=x+4\)

ĐK: x khác 0
Đặt \(x+\frac{1}{x}=a\)\(\Rightarrow\left(x+\frac{1}{x}\right)^2=a^2\Leftrightarrow a^2=x^2+\frac{1}{x^2}+2\cdot x\cdot\frac{1}{x}\Leftrightarrow a^2-2=x^2+\frac{1}{x^2}\)

Có:

\(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2\)
\(=8a^2+4\left(a^2-2\right)^2-4\left(a^2-2\right)a^2\)
\(=8a^2+4\left(a^4-4a^2+4\right)-4\left(a^4-2a^2\right)\)
\(=8a^2+4a^4-16a^2+16-4a^4+8a^2=16\)

Thay \(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=16\)

  vào phương trình, ta có:  \(\left(x-4\right)^2=16\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=-4\\x-4=4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=8\end{cases}}\)Mà điều kiện x khác 0 nên x=8

Vậy phương trình có nghiệm x=8