`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.

are you kidding me?

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

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

\(=2\sqrt{2}\)

a. ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow -5x-5\sqrt{x}+12\sqrt{x}+12=0$

$\Leftrightarrow -5\sqrt{x}(\sqrt{x}+1)+12(\sqrt{x}+1)=0$

$\Leftrightarrow (\sqrt{x}+1)(12-5\sqrt{x})=0$

Dễ thấy $\sqrt{x}+1>1$ với mọi $x\geq 0$ nên $12-5\sqrt{x}=0$

$\Leftrightarrow \sqrt{x}=\frac{12}{5}$

$\Leftrightarrow x=5,76$ (thỏa mãn)

d. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \sqrt{49}.\sqrt{x-2}-14\sqrt{\frac{1}{49}}\sqrt{x-2}=3\sqrt{x-2}+8$

$\Leftrightarrow 7\sqrt{x-2}-2\sqrt{x-2}=3\sqrt{x-2}+8$

$\Leftrightarrow 2\sqrt{x-2}=8$

$\Leftrightarrow \sqrt{x-2}=4$

$\Leftrightarrow x=4^2+2=18$ (tm)

b. ĐKXĐ: $x^2\geq 5$

PT $\Leftrightarrow \frac{1}{3}\sqrt{4}.\sqrt{x^2-5}+2\sqrt{\frac{1}{9}}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$

$\Leftrightarrow \frac{2}{3}\sqrt{x^2-5}+\frac{2}{3}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$

$\Leftrightarrow -\frac{5}{3}\sqrt{x^2-5}=0$

$\Leftrightarrow \sqrt{x^2-5}=0$

$\Leftrightarrow x=\pm \sqrt{5}$

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 \(M=x^3-6x=4^{^{ }3}-6.4=40\)

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 \(M=x^3-6x=4^3-6.4=40\)