Question

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

Answer 1

\(\Leftrightarrow8\left(x^2+\frac{1}{x^2}+2\right)+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x^2+\frac{1}{x^2}+2\right)=x^2+8x+16\) 

Đặt \(a=x^2+\frac{1}{x^2}\) 

PTTT \(8\left(a+2\right)+4a^2-4a\left(a+2\right)=x^2+8x+16\)

\(\Leftrightarrow8a+16+4a^2-4a^2-8a=x^2+8x+16\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-8\end{cases}}\)

Answer 2

đặt ẩn nhé cậu

Answer 3

lam the nao vay

Answer 4

Đặt x+1/x=a

x^2+1/x^2=b 

sau đó làm bình thường

Answer 5

x = 0 được sao?