Encuentra el valor de suelo de un surd continuo finito

Question

Determinado $x=20062007$ y que $$A=\sqrt{x^2+\sqrt{4x^2+\sqrt{16x^2+\sqrt{100x^2+39x+\sqrt{3}}}}}.

Encontrar el mayor entero no exceda de $A$.

Answer 1

$$10x+1<\sqrt{100x^2+39x+\sqrt{3}}<10x+2$$$ $4x+1<\sqrt{16x^2+10x+1}<\sqrt{16x^2+\sqrt{100x^2+39x+\sqrt{3}}}<\sqrt{16x^2+10x+2}<4x+2$ $$$2x+1=\sqrt{4x^2+4x+1}<\sqrt{4x^2+\sqrt{16x^2+\sqrt{100x^2+39x+\sqrt{3}}}}<\sqrt{4x^2+4x+2}<2x+2$$$ $x+1=\sqrt{x^2+2x+1}<A<\sqrt{x^2+2x+2}<x+2$ $

Así$\lfloor A \rfloor=x+1=20062008$.