Question

Tìm x thoả mãn: \(x^4-20x^2+64=0\)

Answer 1

\(\Leftrightarrow\left(x^4-20x^2+100\right)-36=0\)

\(\Leftrightarrow\left(x^2-10\right)^2=36\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-10=6\\x^2-10=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=16\\x^2=4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\pm4\\x=\pm2\end{matrix}\right.\)

 

Answer 2

\(x^4-20x^2+64=0\)

Đặt \(t=x^2\)

\(PT\Leftrightarrow t^2-20t+64=0\\ \Leftrightarrow t^2-16t-4t+64=0\\ \Leftrightarrow t\left(t-16\right)-4\left(t-16\right)=0\\ \Leftrightarrow\left(t-16\right)\left(t-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t-16=0\\t-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}t=16\\t=4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2=16\\x^2=4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\pm\sqrt{16}\\x\pm\sqrt{4}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\pm4\\x=\pm2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\x=2\\x=-2\end{matrix}\right.\\ Vậyx\in\left\{4;-4;2;-2\right\}\)