`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`

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

Ta có x³ = 40 + 6x 

=> M = 40

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

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

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

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

\(=40+6x\)

=>  \(x^3-6x=40\)

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

\(=40+3\sqrt[3]{20^2-14\sqrt{2}^2}.x\)x này là đề bài cho nên thay vào nha bạn

\(=40+3.2.x\)\(hay\)\(x^3=6x+40\Leftrightarrow x^3-6x=40\)(đây là kết quả cần tìm)

x^3 = 40 - 3\(\sqrt[3]{20^2-14^2.2}\).x = 40 - 6x

Tu do tim dc x toi gian hon

are you kidding me?

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

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

\(=2\sqrt{2}\)

bấm máy ra =4

tính máy thì vứt.