Question

Answer 1

Lời giải:
a.

\(\sqrt[3]{(\sqrt{2}+1)(3+2\sqrt{2})}=\sqrt[3]{(\sqrt{2}+1)(\sqrt{2}+1)^2}=\sqrt[3]{(\sqrt{2}+1)^3}=\sqrt{2}+1\)

b.

\(\sqrt[3]{(4-2\sqrt{3})(\sqrt{3}-1)}=\sqrt[3]{(\sqrt{3}-1)^2(\sqrt{3}-1)}\)

\(=\sqrt[3]{(\sqrt{3}-1)^3}=\sqrt{3}-1\)

c.

\((\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4})(\sqrt[3]{3}+\sqrt[3]{2})=(\sqrt[3]{3})^3+(\sqrt[3]{2})^3=3+2=5\)

d.

\(D=\underbrace{\sqrt[3]{4+\frac{5}{3}\sqrt{\frac{31}{3}}}}_{a}+\underbrace{\sqrt[3]{4-\frac{5}{3}\sqrt{\frac{31}{3}}}}_{b}\)

\(a^3+b^3=8\)

\(ab=\sqrt[3]{4^2-(\frac{5}{3}\sqrt{\frac{31}{3}})^2}=\frac{-7}{3}\)

\(D^3=(a+b)^3=a^3+b^3+3ab(a+b)=8-7D\)

\(D^3+7D-8=0\Leftrightarrow (D-1)(D^2+D+8)=0\)

Dễ thấy $D^2+D+8\neq 0$ nên $D-1=0\Leftrightarrow D=1$