Question

Answer 1

Thay \(x=4+2\sqrt{3}\) vào A, ta được:

\(A=\dfrac{\sqrt{3}+1-1}{\sqrt{3}+1+2}=\dfrac{\sqrt{3}}{3+\sqrt{3}}=\dfrac{\sqrt{3}-1}{2}\)

Answer 2

\(x=4+2\sqrt{3}\Leftrightarrow x=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{x}=\sqrt{3}+1\)

Thay \(\sqrt{x}=\sqrt{3}+1\) vào bt A ta có:

\(A=\dfrac{\sqrt{3}-1+1}{\sqrt{3}+2}=\dfrac{\sqrt{3}}{\sqrt{3}+2}\)

Answer 3

\(x=4+2\sqrt{3}\) => \(\sqrt{x}=\sqrt{3}+1\)

thay vào B ta đc

\(\dfrac{\sqrt{3}+4}{\sqrt{3}+2}+\dfrac{5}{\sqrt{3}}+\dfrac{4}{3+2\sqrt{3}}\) = \(\dfrac{3+7\sqrt{3}}{3}\)