Question

[Ngoc An Pham]

Chứng mỉnh rằng

\(\sqrt{5\sqrt{5\sqrt{5...\sqrt{5\sqrt{5}}}}}+\sqrt{6+\sqrt{6+\sqrt{6+...+\sqrt{6+\sqrt{6}}}}}< 8\)

Answer 1

[Chí Cường]

\(\sqrt{5\sqrt{5\sqrt{5...\sqrt{5\sqrt{5}}}}}=x\Rightarrow x^2=5x\Rightarrow x=5\)(n số 5)

Vậy \(\sqrt{5\sqrt{5\sqrt{5...\sqrt{5\sqrt{5}}}}}=5\) khi \(n\rightarrow\infty\)

\(\sqrt{6+\sqrt{6+\sqrt{6+...+\sqrt{6+\sqrt{6}}}}}=x\\ \Leftrightarrow x^2=6+\sqrt{6+\sqrt{6+\sqrt{6+...+\sqrt{6+\sqrt{6}}}}}\\ \Leftrightarrow x^2=6+x\Rightarrow x=3\)(n số 6)

\(\sqrt{6+\sqrt{6+\sqrt{6+...+\sqrt{6+\sqrt{6}}}}}=3\) khi \(n\rightarrow\infty\)

Vậy S < 8