Đặt \(a=\sqrt[3]{7+5\sqrt{2}},b=\sqrt[3]{7-5\sqrt{2}}\)
\(\Rightarrow\left\{{}\begin{matrix}a^3+b^3=14\\ab=-1\end{matrix}\right.\)
Ta có: \(x=a+b+2\Leftrightarrow x-2=a+b\Leftrightarrow\left(x-2\right)^3=\left(a+b\right)^3\Leftrightarrow x^3-6x^2+12x-8=a^3+b^3+3ab\left(a+b\right)\Leftrightarrow x^3-6x^2+12x-8=14+3\left(-1\right)\left(x-2\right)\Leftrightarrow x^3-6x^2+15x-28=0\Leftrightarrow\left(x-4\right)\left(x^2-2x+7\right)=0\Leftrightarrow x-4=0\Leftrightarrow x=4\)
(do \(x^2-2x+7=\left(x-1\right)^2+6\ge6>0\))
Ta có: \(x=\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}+2\)
\(\Leftrightarrow x=\sqrt{2}+1+\sqrt{2}-1+2\)
\(\Leftrightarrow x=2+\sqrt{2}\)
