Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có: \(f\left(x\right)=\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\)= \(\frac{\left(\sqrt{x+1}+\sqrt{x-1}\right)\left(\sqrt{x+1}+\sqrt{x-1}\right)}{\left(\sqrt{x+1}-\sqrt{x-1}\right)\left(\sqrt{x+1}+\sqrt{x-1}\right)}\)=\(\frac{\left(\sqrt{x+1}+\sqrt{x-1}\right)^2}{x+1-\left(x-1\right)}\)
= \(\frac{x+1+x-1+2\sqrt{\left(x-1\right)\left(x+1\right)}}{2}\)= \(\frac{2x+2\sqrt{x^2-1}}{2}\)=\(x+\sqrt{x^2-1}\)
Với a= \(\sqrt{3}\)=> \(f\left(\sqrt{3}\right)=\sqrt{3}+\sqrt{\left(\sqrt{3}\right)^2-1}\)=\(\sqrt{3}+\sqrt{2}\)
\(f\left(x\right)=\frac{2+\sqrt{4+4x}}{2x+2}+\frac{2+\sqrt{4-4x}}{2x-2}\)
\(\Rightarrow f\left(\frac{\sqrt{3}}{2}\right)=\frac{2+\sqrt{4+2\sqrt{3}}}{\sqrt{3}+2}+\frac{2+\sqrt{4-2\sqrt{3}}}{\sqrt{3}-2}\)
\(=\frac{2+\sqrt{3}+1}{\sqrt{3}+2}+\frac{2+\sqrt{3}-1}{\sqrt{3}-2}=\frac{\left(\sqrt{3}+3\right)\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}-\frac{\left(\sqrt{3}+1\right)\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=3-\sqrt{3}-3\sqrt{3}-5=-2-4\sqrt{3}\)