Question

\(\left(\sqrt{\left(\sqrt{20}\right)-\sqrt{19}}\right)^x+\left(\sqrt{\left(\sqrt{20}\right)+\sqrt{19}}\right)^x=2\)

Answer 1

Lời giải:

Đặt \(\sqrt{\sqrt{20}+\sqrt{19}}=a\)

Ta thấy:

\(\sqrt{\sqrt{20}+\sqrt{19}}.\sqrt{\sqrt{20}-\sqrt{19}}=\sqrt{(\sqrt{20}-\sqrt{19})(\sqrt{20}+\sqrt{19})}=\sqrt{20-19}=1\)

\(\Rightarrow \sqrt{\sqrt{20}-\sqrt{19}}=\frac{1}{\sqrt{\sqrt{20}+\sqrt{19}}}=\frac{1}{a}\)

PT trở thành:

\(a^x+\frac{1}{a^x}=2\)

\(\Leftrightarrow a^{2x}+1-2a^{x}=0\)

\(\Leftrightarrow (a^x-1)^2=0\Rightarrow a^x=1\)

Mà \(a\neq 1\) nên \(\Rightarrow x=0\)