Question

D=\(\sqrt[3]{2+10\sqrt{\dfrac{1}{27}}}+\sqrt[3]{2-10\sqrt{\dfrac{1}{27}}}\)

Answer 1

Lời giải:

Đặt \(\sqrt[3]{2+10\sqrt{\frac{1}{27}}}=a; \sqrt[3]{2-10\sqrt{\frac{1}{27}}}=b\)

Khi đó: \(D=a+b\)

\(\left\{\begin{matrix} a^3+b^3=2+10\sqrt{\frac{1}{27}}+2-10\sqrt{\frac{1}{27}}=4\\ ab=\sqrt[3]{(2+10\sqrt{\frac{1}{27}})(2-10\sqrt{\frac{1}{27}})}=\frac{2}{3}\end{matrix}\right.\)

Có:

\(a^3+b^3=(a+b)^3-3ab(a+b)\)

\(\Leftrightarrow 4=D^3-3.\frac{2}{3}D\)

\(\Leftrightarrow D^3-2D-4=0\)

\(\Leftrightarrow (D-2)(D^2+2D+2)=0\)

Vì \(D^2+2D+2=(D+1)^2+1\neq 0\), do đó \(D-2=0\Leftrightarrow D=2\)