Question

Chứng minh rằng với a>\(\dfrac{1}{8}\) thì x=\(\sqrt[3]{a+\dfrac{a+1}{3}\cdot\sqrt{\dfrac{8a-1}{3}}}\)+\(\sqrt[3]{a-\dfrac{a+1}{3}\cdot\sqrt{\dfrac{8a-1}{3}}}\)là số nguyên

Answer 1

Xét \(x^3=2a+3x.\sqrt[3]{a^2-\left(\dfrac{a+1}{3}\right)^2.\dfrac{8a-1}{3}}\)

\(\Leftrightarrow x^3=2a+3x.\sqrt[3]{\dfrac{\left(1-2a\right)^3}{27}}\)

\(\Leftrightarrow x^3=2a+x.\left(1-2a\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+2a\right)=0\)

Dễ thấy \(x^2+x+2a=\left(x+\dfrac{1}{2}\right)^2+\dfrac{8a-1}{4}>0\) (vì \(a>\dfrac{1}{8}\))

Nên x=1 hay x là số nguyên.