Question

Cho \(a=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}\)

Chứng minh rằng: \(\dfrac{64}{\left(a^2-3\right)^3}-3a\) có giá trị là số nguyên

Answer 1

\(a>0\)

Có \(a^3=2-\sqrt{3}+3\sqrt[3]{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\left(\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2-\sqrt{3}}\right)+2+\sqrt{3}\)

\(\Leftrightarrow a^3=4+3a\) 

\(\Leftrightarrow a\left(a^2-3\right)=4\)\(\Leftrightarrow a^2-3=\dfrac{4}{a}\)

\(\Leftrightarrow\dfrac{64}{\left(a^2-3\right)^3}=a^{.3}\) 

\(\Leftrightarrow\dfrac{64}{\left(a^2-3\right)^3}-3a=a^2-3a=4\) là số nguyên.