Đặt \(x=\sqrt[3]{20+14\sqrt[]{2}}+\sqrt[3]{20-14\sqrt[]{2}}\)

\(\Rightarrow x^3=40+3\sqrt[3]{\left(20+14\sqrt[]{2}\right)\left(20-14\sqrt[]{2}\right)}.\left(\sqrt[3]{20+14\sqrt[]{2}}+\sqrt[3]{20-14\sqrt[]{2}}\right)\)

\(\Rightarrow x^3=40+6x\)

\(\Rightarrow x^3-6x-40=0\)

\(\Rightarrow\left(x-4\right)\left(x^2+4x+10\right)=0\)

\(\Rightarrow x=4\)

Vậy \(\sqrt[3]{20+14\sqrt[]{2}}+\sqrt[3]{20-14\sqrt[]{2}}=4\)

em cảm ơn ạ

A = \(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\)

=> A³ = 40 + 6A

<=> A = 4

a) Ta có: \(A^3=\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)^3\)

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

\(=4-3\cdot A\)

\(\Leftrightarrow A^3+3A-4=0\)

\(\Leftrightarrow A^3-A+4A-4=0\)

\(\Leftrightarrow A\left(A-1\right)\left(A+1\right)+4\left(A-1\right)=0\)

\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)

\(\Leftrightarrow A=1\)

`x=root{3}{14sqrt2+20}+sqrt{-14sqrt2+20}`

`<=>x^3=14sqrt2+20-14sqrt2+20+3root{3}{(14sqrt2+20)(20-14sqrt2)}(root{3}{14sqrt2+20}+sqrt{-14sqrt2+20})`

`<=>x^3=40+3root{3}{400-392}.x`

`<=>x^3=40+6x`

`<=>x^3-6x=40`

\(x=\sqrt[3]{30+14\sqrt{2}}-\sqrt[3]{20+14\sqrt{2}}\)

\(=\sqrt[3]{\left[2^3+3.2^2.\sqrt{2}+3.2+\sqrt{2^2}+\left(\sqrt{2}\right)^3\right]}+\sqrt[3]{\left[2^3-3.2.\sqrt{2}+3.2.\sqrt{2^2}-\left(\sqrt{2}\right)^3\right]}\)

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

\(=2+\sqrt{2}+2-\sqrt{2}\)

\(=4\)

Vậy x = 4.

Ta có: \(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\)

\(=\sqrt[3]{8+12\sqrt{2}+12+2\sqrt{2}}+\sqrt[3]{8-12\sqrt{2}+12-2\sqrt{2}}\)

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

\(=2+\sqrt{2}+2-\sqrt{2}\)

\(=4\)