Find an integer most close to A

[Albert1]

$A=\sqrt{\dfrac{1}{\sqrt[3]9-2}+2\sqrt[3]9}$

find an integer $B$ most close to A

 

[anemone]

My solution:

Spoiler

$\begin{align*}A&=\sqrt{\dfrac{1}{\sqrt[3]9-2}+2\sqrt[3]9}\\&=\sqrt{\dfrac{1(9^{\frac{2}{3}}+2(9^{\frac{1}{3}})+2^2)}{(9^{\frac{1}{3}}-2)(9^{\frac{2}{3}}+2(9^{\frac{1}{3}})+2^2)}+2\sqrt[3]9}\\&=\sqrt{\left(\dfrac{9^{\frac{2}{3}}+2(9^{\frac{1}{3}})+2^2}{1}\right)+2(9^{\frac{1}{3}})}\\&=\sqrt{9^{\frac{2}{3}}+4(9^{\frac{1}{3}})+2^2}\\&=\sqrt{(9^{\frac{1}{3}}+2)^2}\\&=9^{\frac{1}{3}}+2\end{align*}$

Note that

$2^{3\left(\frac{1}{3}\right)}+2<9^{\frac{1}{3}}+2<3^{3\left(\frac{1}{3}\right)}+2$

We get:

$4<A<5$

Therefore $B=4$.

 

[Albert1]

If you said :$B=4$ then you should say :$4<A<4.5$
then $B=4$ is closer to $A$

 

[anemone]

If you said :$B=4$ then you should say :$4<A<4.5$
then $B=4$ is closer to $A$

Ops...my bad...I thought

Spoiler

the question asked about $\lfloor A\rfloor=B$.