Question

Chứng minh : \(\frac{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}< \frac{1}{3}\)

 

Answer 1

đặt \(\sqrt{2+\sqrt{2+\sqrt{2}}}=a\)

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

\(=\frac{1}{2+\sqrt{2+x}}\)

ta có \(2+x>2\)

=>\(\sqrt{2+x}>\sqrt{2}\)

=>2+\(\sqrt{2+x}>2+\sqrt{2}\)

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

=>\(\frac{1}{2+\sqrt{2+x}}< \frac{1}{3}\)