Question

\(\frac{\sqrt{2x+2\sqrt{x^2-4}}}{\sqrt{x^2-4}+x+2}\)

cho \(x=2\left(\sqrt{3}+1\right)\)

rút gọn và tính giá trị biểu thức

Answer 1

\(=\frac{\sqrt{x+2+2\sqrt{\left(x+2\right)\left(x-2\right)}+x-2}}{\sqrt{\left(x-2\right)\left(x+2\right)}+x+2}\)

\(=\frac{\sqrt{\left(\sqrt{x+2}+\sqrt{x-2}\right)^2}}{\sqrt{x+2}.\left(\sqrt{x+2}+\sqrt{x-2}\right)}\)

\(=\frac{\sqrt{x+2}+\sqrt{x-2}}{\sqrt{x+2}.\left(\sqrt{x+2}+\sqrt{x-2}\right)}=\frac{1}{\sqrt{x+2}}\)

Thay \(x=2+2\sqrt{3}\)

\(=\frac{1}{\sqrt{4+2\sqrt{3}}}=\frac{1}{\sqrt{\left(\sqrt{3}+1\right)^2}}=\frac{1}{\sqrt{3}+1}\)

\(=\frac{\sqrt{3}-1}{2}\)