Question

[Mynnie]

Giải giúp mình bài này với ạ!
\(\sqrt[3]{2+\sqrt[]{5}}+\sqrt[3]{2-\sqrt[]{5}}\)
 

Answer 1

[Trần Bảo Như]

Đặt \(A=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\)

Áp dụng \(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\) ta có:

\(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\)

\(A^3=2+\sqrt{5}+2-\sqrt{5}+3\sqrt[3]{4-5}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)

\(=4-3A\)

Giải PT:

\(A^3+3A-4=0\Leftrightarrow A^3-1+3A-3=0\)\(\Leftrightarrow\left(A-1\right)\left(A^2+A+1\right)+3\left(A-1\right)=0\)\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}A-1=0\\A^2+A+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}A=1\\A^2+2.\frac{1}{2}A+\frac{1}{4}-\frac{1}{4}+4=0\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}A=1\\\left(A+\frac{1}{2}\right)^2+\frac{15}{4}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}A=1\\\left(A+\frac{1}{2}\right)^2=-\frac{15}{4}\left(L\right)\end{cases}}}\)

Vậy \(A=1\)