Question

[Nguyễn Hữu Tâm]

Giải phương trình sau:                                           

Answer 1

[Nguyễn Đăng Nhân]

Đặt \(x+\frac{1}{x}=a\), ta có:

\(\left(x+\frac{1}{x}\right)^2=a^2\Leftrightarrow x^2+2+\frac{1}{x^2}=a^2\Rightarrow x^2+\frac{1}{x^2}=a^2-2\)

\(\Rightarrow8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=\left(x+4\right)^2\)

\(\Leftrightarrow8a^2+4\left(a^2-2\right)^2-4\left(a^2-2\right)^2a^2=\left(x+4\right)^2\)

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

\(\Leftrightarrow8a^2+4a^4-16a^2+16-4a^4+8a^2-\left(x+4\right)^2=0\)

\(\Leftrightarrow\left(x+4\right)^2=16\)

\(\Rightarrow\hept{\begin{cases}x=0\\x=-8\end{cases}}\)

Answer 2

[Đoàn Duy Nhật]

x=0 hoặc x=-8 nha

~HT~

 mình đang tăng SP mong bn k cho mình nhé

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