Question

[Vương Thiên Nhi]

chứng minh rằng với \(a>\frac{1}{8}\) thì số sau đây là một số nguyên dương.

\(x=\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}+\sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}\).

Answer 1

[Nguyễn Việt Lâm]

\(x^3=2a+3x.\sqrt[3]{a^2-\frac{\left(a+1\right)^2}{9}\left(\frac{8a-1}{3}\right)}\)

\(x^3=2a+3x\sqrt[3]{\frac{1-6a+12a^2-8a^3}{27}}\)

\(x^3=2a+3x\sqrt[3]{\left(\frac{1-2a}{3}\right)^3}\)

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

\(x^3-x+2ax-2a=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=1\\x^2+x+2a=0\left(1\right)\end{matrix}\right.\)

Xét (1): \(x^2+x+\frac{1}{4}+2a-\frac{1}{4}=0\Rightarrow\left(x+\frac{1}{2}\right)^2+2\left(a-\frac{1}{8}\right)=0\)

Do \(a>\frac{1}{8}\Rightarrow\left(x+\frac{1}{2}\right)^2+2\left(a-\frac{1}{8}\right)>0\)

\(\Rightarrow\left(1\right)\) vô nghiệm \(\Rightarrow x=1\) hay x nguyên dương