Question

Cho \(x=\frac{1}{2}\left(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\right)\) với a>0; b>0

Tính \(B=\frac{2\sqrt{x^2-1}}{x-\sqrt{x^2-1}}\)

Answer 1

Lời giải:

\(2x=\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\)

\(\Rightarrow 4x^2=\frac{a}{b}+\frac{b}{a}+2\Rightarrow 4(x^2-1)=\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2\)

\(\Rightarrow \sqrt{4(x^2-1)}=|\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}|\)

Nếu \(a\geq b\Rightarrow \sqrt{4(x^2-1)}=\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\)

Do đó: \(\frac{B}{2}=\frac{\sqrt{4(x^2-1)}}{2x-\sqrt{4(x^2-1)}}=\frac{\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}}{\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}-\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}\)

\(\Rightarrow B=\frac{a-b}{b}\)

Tương tự với \(a< b\)