Question

[Thi Bùi]

cho x^2-5x+1=0 tinh gia tri x^2+1/x^2

Answer 1

[Nguyễn Minh Đăng]

Ta có: \(x^2-5x+1=0\)

\(\Leftrightarrow\left(x^2-5x+\frac{25}{4}\right)-\frac{21}{4}=0\)

\(\Leftrightarrow\left(x-\frac{5}{2}\right)^2=\frac{21}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{5}{2}=\frac{\sqrt{21}}{2}\\x-\frac{5}{2}=-\frac{\sqrt{21}}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5+\sqrt{21}}{2}\\x=\frac{5-\sqrt{21}}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2=\frac{26+10\sqrt{21}}{4}\\x^2=\frac{26-10\sqrt{21}}{4}\end{cases}}\)

Thay vào ta được: \(\frac{x^2+1}{x^2}=1+\frac{1}{x^2}\) đến đây: \(\orbr{\begin{cases}=1+\frac{1}{\frac{26+10\sqrt{21}}{4}}\\=1+\frac{1}{\frac{26-10\sqrt{21}}{4}}\end{cases}}\)

\(\orbr{\begin{cases}=1+\frac{4}{26+10\sqrt{21}}=\frac{30+10\sqrt{21}}{26+10\sqrt{21}}\\=1+\frac{4}{26-10\sqrt{21}}=\frac{30-10\sqrt{21}}{26-10\sqrt{21}}\end{cases}}\)