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

Đặt \(\sqrt[3]{20+14\sqrt{2}}=a;\sqrt[3]{20-14\sqrt{2}}=b\).Từ đó => a + b = x và ab=2

\(\Rightarrow x^3=40+3ab\left(a+b\right)\)

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

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

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

Dễ thấy \(x^2+4x+10=\left(x+2\right)^2+6>0\)

\(\Rightarrow x=4\).Thay vào ta tìm được P = 1969

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

Ta có x³ = 40 + 6x 

=> M = 40

Lời giải:

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

\(\Rightarrow \left\{\begin{matrix} a^3+b^3=40\\ ab=\sqrt[3]{(20+14\sqrt{2})(20-14\sqrt{2})}=\sqrt[3]{20^2-(14\sqrt{2})^2}=2\end{matrix}\right.\)

Do đó:

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

\(\Leftrightarrow x^3=40+3.2.x\)

\(\Leftrightarrow x^3-6x-40=0\Leftrightarrow x^2(x-4)+4x(x-4)+10(x-4)=0\)

\(\Leftrightarrow (x^2+4x+10)(x-4)=0\)

\(\Rightarrow x-4=0\Rightarrow x=4\) (do $x^2+4x+10>0$)

Vậy \(M=x^3-6x=4^3-6.4=40\)

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

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

\(\Leftrightarrow x^3-6x=40\)

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

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

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

= 4

Thay x=4 vào biểu thức \(M=x^3-6x=4^{^{ }3}-6.4=40\)

`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`

b. ĐK \(\hept{\begin{cases}x-2\ge0\\y+2014\ge0\\z-2015\ge o\end{cases}\Rightarrow\hept{\begin{cases}x\ge2\\y\ge-2014\\z\ge2015\end{cases}}}\)

Ta có \(\sqrt{x-2}+\sqrt{y+2014}+\sqrt{z-2015}=\frac{1}{2}\left(x+y+z\right)\)

Đặt  \(\hept{\begin{cases}\sqrt{x-2}=a\ge0\\\sqrt{y+2014}=b\ge0\\\sqrt{z-2015}=c\ge0\end{cases}}\Rightarrow\hept{\begin{cases}x-2=a^2\\y+2014=b^2\\z-2015=c^2\end{cases}\Rightarrow x+y+z}=a^2+b^2+c^2+3\)

Pt \(\Leftrightarrow a+b+c=\frac{1}{2}\left(a^2+b^2+c^2+3\right)\Leftrightarrow a^2+b^2+c^2+3=2a+2b+2c\)

\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\Leftrightarrow\hept{\begin{cases}a-1=0\\b-1=0\\c-1=0\end{cases}}\)\(\Leftrightarrow a=b=c=1\)

\(\Rightarrow\hept{\begin{cases}x-2=1\\y+2014=1\\z-2015=1\end{cases}\Rightarrow\hept{\begin{cases}x=3\\y=-2013\\z=2016\end{cases}\left(tm\right)}}\)

Vậy \(x=3;y=-2013;z=2016\)